Does spin add energy to a quantum mechanical system? If a particle (let's say an electron, a 1/2 spin particle) has spin component $S_z=\hbar/2$. Does the spin contribute to the total energy of the electron? 
For example, without considering the spin, a particle in a harmonic oscillator potential has energy $(n+1/2)\hbar \omega$. When we consider the spin, should we consider any kind of extra energy?
My reasoning is that there is some energy attributed to orbital angular momentum and spin is angular momentum. Or is this assumption wrong?
 A: It's impossible to say in the framework of existing physical theory if "spin" is part of an electron's energy.
The reason for this is that, in order to say some part of something does/doesn't "contribute" something to the whole, you have to have a sense of the "other parts" as separable therefrom, as a something "missing" that part, against which you can make the comparison. In that case, that means you would need to have a sense of an "electron without spin", to compare the energy of against electrons with spin, and find the energy difference, which will be the total contribution due to spin.
But there's a problem. There are no electrons without spin. The whole point of the spin is that it is an intrinsic angular momentum that can never be taken away nor added to. It is part of what makes the electron what it is. Therefore, the necessary point of comparison doesn't exist. Just as with charge: there are no electrons without charge, so asking how much the charge influences things, how much "energy comes from having charge", which would then necessitate comparison to some putative "chargeless" version, is similarly meaningless.
Of course, you can posit some theory in which there is an object called a "spinless electron", but then it would depend entirely on the newly-posited details, and there is no evidence anything that can be reasonably thought of as such actually exists in our real-life world. Hence any answer would have to be entirely fictitious.
With regard to your question about total system energy, however, that is much simpler to answer. There is a key observation that one should make regarding the total energy of an isolated mechanical system like this in the framework of theory:

It is, in effect, an arbitrary quantity.

The key point rather, is that that quantity does not change, i.e. it is conserved. In any mechanical system, classical or quantum, you have a total mechanical energy of
$$E_\mathrm{mech} = K + U$$
This is always the case. The only difference with quantum mechanics is that, like all physical parameters, the amount of information defining what it is may be restricted, which shows up by requiring us to write little hats on top of each term there. Regardless, here's the thing: you may have heard that the potential energy, $U$, can be altered by an arbitrary constant. (If you haven't, here goes, and you'd better.) But it goes further. Note that that means that the total $E_\mathrm{mech}$ can likewise be altered by such, and we can hence consider adding such a constant independent of either as well - it doesn't matter mathematically, as it's just a sum. That is, there is nothing wrong at all - nothing within the framework of the physical theory for that system that breaks - if you add to an $E_\mathrm{mech}$ some completely arbitrary constant term, that never affects any dynamics, and not associated with any of the others:
$$E_\mathrm{mech}' = K + U + \mbox{arbitrary constant}$$
This is sometimes phrase as saying that only differences in energy are physically meaningful.
And as per the above discussion, since electron spin cannot change, if it does in some sense "contribute some energy", that energy has to serve as an arbitrary constant in the context of this and any discussion of an electron system And, just as we can add such constants, we can also elide them, with no consequence.
ADD: I want to also discuss something. Since spin is the only quantum mechanical quantity that makes any sense to call as the rotational angular momentum of the electron, you might be tempted to use the formula
$$K_\mathrm{rotation} = \frac{1}{2} I\omega^2 = \frac{L^2}{2I}$$
from Newtonian mechanics, which can be turned to a quantum equation by replacing $\omega$ with a quantum operator. The problem here is the electron moment of inertia $I$ about its axis must be zero, because it is a point - since $I = k \cdot m_e r^2$, but electron radius $r = 0$ at least according to current theory. But that gives a zero divisor for finite angular momentum $L$ (here $\frac{\sqrt{3}}{2} \hbar$) and infinite kinetic energy, which is similar to the case for a massless particle and linear kinetic energy - you get a similar infinity at finite linear momentum and need special relativity to get $K_\mathrm{translation} = \frac{p}{c}$, but I don't believe there's an analogous relativistic expression for rotational kinetic energy in the zero rotational inertia regime (c.f. Ehrenfest paradox). I could be wrong, though! That might provide some way to define a quantity you could call "spin energy of the electron", but I don't know.
A: There is nothing rotating in the electron which is assigned spin 1/2, so in this sense there is no energy associated with spin the way that there is energy associated with angular momentum.
On the other hand, spin plays a role in bound states, as for example, in the bound states of hydrogen 


The small splitting of the spectral line is attributed to an interaction between the electron spin S and the orbital angular momentum L. It is called the spin-orbit interaction. 

So the existence of spin does affect the energy behavior of bound matter.
A: In General Relativity, spin do adds something to the energy.  Consider a Kerr-Newman black hole, of total mass $M$, spin $S$ and charge $Q$.  These parameters are linked into a well known relation (remember that mass and energy are related, in relativity theory: $E = M c^2$):
\begin{equation}\tag{1}
M = \sqrt{\Big( M_0 + \frac{k \, Q^2}{4 G M_0} \Big)^2 + \Big( \frac{S c}{2 G M_0} \Big)^2},
\end{equation}
where $M_0$ is the irreducible mass and $k$ is the vacuum dielectric constant.  According to this formula, spin adds a contribution to the BH total mass, like charge (the energy stored in the electric field adds a contribution to the total mass).
Since a BH doesn't have "parts" (i.e. it isn't made of material particles), it is an "abstract object" as much as an electron (which isn't a little rotating ball of matter).  Likewise, a "rotating" (i.e. spinning) BH isn't an ordinary object made of moving "material parts".
Take note however that formula (1) above doesn't apply to an electron.  The numbers doesn't fit!  But it's a way to show that spin could contribute to the total mass of an object, under the frame of General Relativity, because of gravity (there's a $G$ in there!).
