Do all equations have identical units on the left- and right-hand sides? Do all equations have 
$$\text{left hand side unit} = \text{right hand side unit}$$
for example,
$$\text{velocity (m/s)} = \text{distance (m) / time (s)},$$
or is there an equation that has different units on the left- and right-hand sides? I would like to consider 
empirical equations (determined from experimental results) and theoretical equations (derived from basic theory).
 A: It doesn't matter where the equation came from - a fit to experimental data or a deep string theoretic construction - or who made the equation - Albert Einstein or your next-door neighbour - if the dimensions don't agree on the left- and right-hand sides, it's nonsense.
Consider e.g. my new theory that the mass of an electron equals the speed of light. It's just meaningless nonsense from the get-go. 
This isn't that restrictive - there's lots of equations with correct dimensions (though in some cases you can derive equations or estimates by so-called dimensional analysis, where you just make sure the units agree).  But it is useful for checking your work. If you derive a result and the dimensions don't agree, you know you must have made a mistake.
There is a subtle distinction between unit and dimension. A dimension represents a fundamental quantity - such as mass, length or time - whereas a unit is a man-made measure of a fundamental quantity or a product of them - such as kg,  meters and seconds. Arguably, one can write meaningful equations such as 60 seconds = 1 minute, with matching dimensions but mismatching units (as first noted by Mehrdad).
A: Every equation should have corresponding dimension. Either by the natural dimensions of the equation
$$\text{Average Speed} = \frac{\text{Distance}}{\text{Time}}$$
or by some constant which gives the correct dimension
$$F = \frac{Gm_1 m_2}{r^2}$$
Where $G$ has dimension $[M^{-1}] [L^3] [T^{-2}]$ to ensure that the dimensions are equal on both sides.
A: They have to be equal, because if the units are not identical, we will add fudge factors to make them identical.
What you are looking at is called dimensional analysis.  Dimensional analysis is a tool that lets us turn equations like $x(t) = \frac{1}{2}at^2 + vt + x_0$ into something meaningful in the real world.
The real truth is that there are no "units" in the physical world.  Units are a concept that was created by humans because we found it useful to model the world in a way that includes units.  They are a way we can compare one value to another in a very objective manner.  The way units were constructed by humans made them as useful as possible.  We noticed that it was possible to view the world where $3(m) + 2(m)$ or even $3(cm) + 2(in)$ lead to meaningful results while $3(m) + 2(s)$ did not yield meaningful results.
There are times where this has had its complications.  For example, Robert Hooke noticed that there was a proportional relationship between the distance you pull a spring and the force it exerts: $F\propto x$.  This relationship has been canonized as Hooke's law in the form $F=kx$.  Now we know that $x$ is a unit of length and $F$ is a unit of force.  We would like to claim that we can do dimensional analysis, so we cleverly choose to quote $k$ in $(\frac{N}{m})$ so that the units line up.
Thus:


*

*When one is in the learning phase of their physics careers, dimensional analysis will always work because the smart people who developed the constants chose the units on their constants to make them work.

*Just because the units line up does not mean the equation is "true."

*Early on, if the units do not line up, you may assume that means the equation is "false," and you'll be right almost all of the time.  If one becomes a physicist, later on there is a chance the units do not line up because you have discovered a new property where no one smart has gotten the units right for you.  In this case, you have the privileged of choosing the units on your constant so that they work out for the next person!

A: There is a simple argument to see why dimensions must agree on both sides. To use innisfree's example, consider the (obviously wrong) equation
$$m_e = c\tag{*}$$
$m_e$ being the mass of the electron and $c$ the speed of light. I assume I have written this equation in the International System units (kilograms, meters, seconds). Now if I want to write this equation in another unit system, for instance (grams, centimeters, seconds). Then I will have a new numerical value of my mass and my speed that reads
$$m_e' = 1000 m_e,$$
$$c'=100 c.$$
Under this transformation, equation (*) becomes
$$m_e'/1000 = c'/100$$
$$\Rightarrow m_e' = 10 c'.$$
I get a different equation than (*). Physics should be the same in any system of units because units are just a human contruction made to measure quantities, it does not impact on how Nature behaves. Thus this equation must be wrong.
Now if you take a correct equation such as Newton's second law
$$F = m\cdot a$$
written in the International system of units and you make the same transformation from (kilograms, meters, seconds) to (grams, centimeters, seconds) and express the new values $F'$, $m'$, $a'$ in this new system of units, you can check that you get a new equation
$F' = m'\cdot a'$
which is the same equation as before. Newton's second law is unit-system invariant, as should any correct equation be.
A: It depends what you mean by "unit".
If you mean something like "seconds", then no.
Counterexample: 1 minute = 60 seconds has different units on both sides, but they're both representing a duration, so they can still be equal.
If you mean something like "time", then yes. An equation means two things are equal, i.e. the same. For that to be true, they have to be the same type of thing. You can't compare people to numbers, you can't compare distance to time, you can't compare temperature to pressure...
A: Yes for sure. After all you cannot say 5 chickens = 2 buffaloes. 
Here is an excerpt from NCERT physics for class 12 Chapter 2. I hope this helps.

The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or subtracted which have the same dimensions.  A thorough understanding of dimensional analysis helps us in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions.  When magnitudes of two or more physical quantities are multiplied, their units should be treated in the same manner as ordinary algebraic symbols. We can cancel identical units in the numerator and denominator.  The same is true for dimensions of a physical quantity.  Similarly, physical quantities represented by symbols on both sides of a mathematical equation must have the same dimensions.

Psst :: Many a times you can just use the dimensional analysis to check for answers in multiple choice questions. 
A: As mentioned by the other answers, it is the dimension which essentially needs to be the same on both sides of an equation and not the unit. This has already been spoken with the example of $1 \,\mathrm{hour} = 60 \,\mathrm{minutes}$.  
Let me give you one example and illustrate to you that:
The thing called unit has been merely developed for our own convenience so that we can avoid all the messy mathematical work with huge numbers and exponents.
In particle physics, for convenience, they consider the speed of light to be unity, or in other words, $c=1$. Yes, that too without any unit! And then you get the equation formulated by Einstein and the most used equation in particle, nuclear and high-energy physics i.e. $E=mc^2$ to become $E=m$. Yup, you can easily notice that in this equation, dimensions are not equal on both sides of the equation. You might also realise that then the units on both sides of the equation will also be different. But here, fortunately they have defined the unit of mass and energy to be the same i.e. $\mathrm{electron-volts\,\ (eV)}$.
And further more, they also express temperature in MeV and GeV instead of Celsius or Kelvin!
The reason for this peculiar way of using units is that particle physicists and high-energy physicists require to convert energy to mass and vice-versa to find out the energy and the temperature required to produce the collision that might generate a certain particle and so on.
Hence it is laborious if they have to use different units all the time while assessing the huge data of collisions. so for the sake of simplicity, they take the speed of light to be $1$ and express mass, energy and temperature using the same unit.
This clearly illustrates that units are for our own convenience. It is up to us how we want to use it, although keeping the first statement of my answer in mind. 
A: The dimensional units in an equation must balance. Sometimes a dimensionless "unit" may appear on one side and not be obvious (or even present) on the other side.  For example, consider the kinetic energy of a spinning object: $$K_s = \frac{1}{2}\mathcal{I}\omega^2.$$
A comparison of SI units yields the following:
$$[J]=[kg\cdot m^2]\frac{[rad]^2}{[s]^2}$$
The right side reduces to $[J][rad]^2$, which seems to be different from the left ([J]), but a radian is dimensionless, so that it doesn't affect the dimensional balance of the equation.  Other rotational equations may have the same effect.
A: No. All equations have the same dimension on both sides. Dimensions are mass, distance, time, speed, acceleration, force, power, electric current, electric charge etc.
As long as you work with symbolic relations, you only care about dimensions. The equation
$$v = \frac{s}{t}$$
(velocity = distance / time) works with any units as long as they are units for the appropriate dimensions. For example the equation
$$17\ \mathrm{knot}\ \dot=\ \frac{2266\ \mathrm{km}}{3\ \mathrm{day}}$$
holds.
However, to evaluate it, you need to convert the units to common base. Obviously, $\frac{2266}{3}$ does not equal $17$. It equals $755.\bar 3$ and then you have to use the appropriate conversion factors for $1\ \mathrm{day} = 24\ \mathrm{h}$ and $1.852\ \mathrm{km}=1\ \mathrm{nmi}$ to get the value in knots.
And you can do it in two ways:
$$\frac{2266 \mathrm{km}}{3\ \mathrm{day}} = \frac{\frac{2266\ \mathrm{km}}{1.852\ \mathrm{km/nmi}}}{3\ \mathrm{day}\cdot 24\ \mathrm{h/day}}\ \dot=\ \frac{1224\ \mathrm{nmi}}{72\ \mathrm{h}}\ \dot=\ 17\ \mathrm{nmi/h}$$
or
$$\frac{2266 \mathrm{km}}{3\ \mathrm{day}}\ \dot=\ 775.3\ \mathrm{km/day}\cdot 24\ \mathrm{h/day}\ \dot=\ \frac{31.47\ \mathrm{km/h}}{1.852\ \mathrm{km/nmi}}\ \dot=\ 17\ \mathrm{nmi/h}$$
Note, that the conversion factors are themselves equations with different units on each side, and they are essential for doing unit conversions.
A: It's worth noting natural unit systems, which may appear to violate this rule.
Since certain physical constants (e.g., $G$ and $c$) simply reflect an arbitrary choice of units, it can be convenient to change units so that they are identically 1.
For example, in Planck units, where $G=c=1$, we can write the Schwarzschild radius as $r_s = 2m$.  While it appears that the dimensions don't match up, this is actually telling us that the Schwarzschild radius of an object (measured in Plank lengths) is twice its mass (measured in Planck masses).
There's even a unique way of reinstating these constants afterwards in order to recover the usual form of the equation.  In this case, we know that the LHS has dimensions of length, and the RHS dimensions of mass, so we must insert a constant of dimensions $[L][M]^{-1}$ on the RHS.  We can construct this as $G/c^2$, and so we end up with the familiar form:
$$
r_s=\frac{G}{c^2}\cdot2m.
$$
Natural units have some unusual consequences, in that any two physical quantities can be (somewhat) meaningfully added and compared, regardless of their dimensions.  For example, the energy–momentum relation of special relativity is written in Planck units as
$$
E^2 = p^2 + m^2.
$$
Similarly, Einstein's familiar energy-mass relation reduces, delightfully, from $E=mc^2$ to simply $E=m$.  Whether you consider this convenient or confusing is up to you!
A: Hint: what do you get if you add two apples to three oranges? You can only add like things to like things. In this sense it is technically correct to add 1 m to 3 inches and quote the result as 1 m 3 inches (both are measurements of lengths), but it would not be very useful or good practice.
There are seven fundamental units: kilogram, metre, candela, second, ampere, kelvin, and mole. All other physical quantities can be derived in terms of these (check out here).
A: Few of the answers thus far have directly addressed empirical equations, such as the following:
Vapor pressure of isobutane
(source)
$$
\log_{10}{P_\mathrm{mmHg}}=6.74808-{882.80\over 240.0+T_\mathrm{^\circ C}}
$$
Seeton model for kinematic viscosity of various liquids
(source; $K_0$ is the zero-order modified Bessel function of the second kind)
$$
\ln{\left\{\ln{\left[\nu_\mathrm{cSt} + 0.7
 + e^{-\nu_\mathrm{cSt}}K_0\!\left(\nu_\mathrm{cSt} +
                                          1.244067\right)\right]}\right\}}
       = A - B\ln{T_\mathrm{^\circ K}}
$$
Density of supercritical carbon dioxide
(source)
$$
\rho_\mathrm{kg\over m^3} = \sum_{i=0}^4{A_iP_\mathrm{psia}^i}
$$
where each $A_i$ is calculated as:
$$
A_i = \sum_{j=0}^4{b_{ij}T_\mathrm{^\circ C}^j}
$$
In all of these equations, the units (as distinct from the dimensions, as emphasized in other answers) that must be be used for each variable are specified as a subscript ("$\mathrm{cSt}$" indicates centiStokes units). As is evident, if these units were to be included explicitly in the calculations it would wreck the dimensionality of all three relationships, including violating the requirement that arguments to transcendental functions must be strictly dimensionless.  
Instead, in equations like these only the numerical values of the variables in the specified units are used.  Effectively, all the quantities are non-dimensionalized by dividing each by the relevant unit.  Thus, for example, to obtain the isobutane vapor pressure at $25~^\circ\mathrm C$, one substitutes ${25~\mathrm{^\circ C} \over \mathrm{^\circ C}}=25$ for $T_\mathrm{^\circ C}$ and obtains $P_\mathrm{mmHg} = 2611 = {2611~\mathrm{mmHg}\over \mathrm{mmHg}}$ as the result, which then is re-dimensionalized to $2611~\mathrm{mmHg}$.
