Since it is a recurrent question, I have rather to add my answer here than to more recent ones.

I hope I can make more clear some points which was not completly well focused in some previous answers.

Kinematic description

Once we have chosen whatever reference frame we like (here does not matter if inertial or not) and we have a description of the trajectories of N bodies, say N vectors ${\bf r}_i(t)$, we can always use a reference frame centered on one of the bodies, say the a-th, by simply subtracting the position vector of the chosen body to any other position vector. Therefore, in this new reference frame the trajectories of the original system of N bodies will be: $$ {\bf r}^{\prime}_i(t) = {\bf r}_i(t) - {\bf r}_a(t).~~~~~~~~~~~~~~~[1] $$ It is clear that in this new frame ${\bf r}^{\prime}_a(t)=0$ by construction, i.e. the a-th body is at rest forever.

An example of such transformation of coordinates is the change of reference frame required if we want to find the proper description of the Solar system as seen by an observer on the Earth, starting from trajectories in the (inertial) reference frame where the center of mass of the Solar system is at rest. Notice that an observer at rest on the surface of Earth is not only translating with the planet, with respect to the center of mass, but she/he is also rotating, so the actual transformation would be more complicate than eqn. [$1$]. However we can ignore the need of an additional rotation of our vectors if we confine our considerations to reference frames which do not rotate relatively to the original frame.

At this point it should be clear that there is nothing wrong to describe the motion of the Solar system bodies from the Earth. It is just one of the infinite possible choices of the origin of the reference frame, an probably the most useful for Earth based observers. It has the same right to be used as a reference frame fixed on a moving car to describe what passengers see.

However, the possibility of changing point of view, does not imply that different choices would provide the same description of the trajectories in an N-body system. Quite interestingly, if we start from a reference frame where body $a$ is at rest, i.e. ${\bf r}_a(t)=0$, where a second body b moves according to ${\bf r}_b(t)$, and we move to a new reference frame based on body $b$, in the new system body $a$ will be described by the vector ${\bf r}^{\prime}_a(t)=-{\bf r}_b(t)$. This implies that motion of $a$ as seen by $b$ or motion of $b$ as seen by $a$ differ only by an inversion and therefore they have the same synthetic description.

What about applying the above consideration to the Earth-Sun system? In the case of the two-body system things are quite simple. The shape of Earth trajectory as seen from Sun or that of Sun as seen from Earth are the same. In addition, since the center of mass of the Sun-Earth system is within Sun, Earth trajectory as seen from Sun is almost coinciding with the same orbit as described from the center of mass.

In the next two figure I have plotted the orbit of the two bodies in the reference frame of the center of mass

and in the reference frame of (non-rotating) Earth. Distance units are millions of kilometers.

Things change a lot when we describe motion of other Solar system bodies. THe next two plots show the motion of Sun, Venus, Earth, Mars and Jupiter, as seen from the center of mass of the system or from (non-rotating) Earth.

Even at this kinematic level, the greater simplicity of the description in the center of mass frame is evident. Nevertheless, I want to stress once more that nothing is wrong with this description. It is the closest to what we get frome Earth-based observations.

Dynamic description

From the point of view of solving a problem of Newtonian dynamics, we all know that the center of mass reference frame of an N-body system is convenient. Since it is an inertial frame, we can use Newton's law ${\bf F}=m{\bf a}$ in connection with Newton's gravitional force law, without need of introducing additional inertial forces.

Notice however that once one has written the set of differential equations of motion for the gravitational N-body problem: $$ {\bf a}_i = G\sum_{j\neq i} m_j \frac{({\bf r}_j - {\bf r}_i)}{\left|{\bf r}_j - {\bf r}_i\right|^3} $$ it is trivial to write down the equations of motion referred to body $a$: $$ {\bf a'}_{i} = G\sum_{j\neq i} m_j \frac{({\bf r'}_j - {\bf r'}_i)}{\left|{\bf r'}_j - {\bf r'}_i\right|^3} - G\sum_{j\neq a} m_j \frac{({\bf r'}_j - {\bf r'}_a)}{\left|{\bf r'}_j - {\bf r'}_a\right|^3} ~~~~~~~~~~~[2] $$ where $$ {\bf r'}_i = {\bf r}_i - {\bf r}_a $$ and $$ {\bf a'}_i = {\bf a}_i - {\bf a}_a = \frac{{\mathrm d}^2 ({\bf r}_i - {\bf r}_a ) }{{\mathrm d}t^2}. $$ There are two interesting things to notice in eqn. 2, the first one could help to clarify some statements present in other answers:

1. the second term in the right hand side is nothing but the contribution of the inertial force to acceleration of particle $i$-th. It is a purely translational term and no centrifugal or Coriolis force appears, since the non-inertial reference frame coinciding with body $a$ is not rotating.
2. in the special case of the two-body problem, eqn.[$2$] for body $1$ becomes: $$ {\bf a'}_{1} = G m_2 \frac{ ({\bf r'}_2 - {\bf r'}_1) }{ \left|{\bf r'}_2 - {\bf r'}_1\right|^3 } - G m_1 \frac{ ({\bf r'}_1 - {\bf r'}_2) }{ \left|{\bf r'}_1 - {\bf r'}_2\right|^3 } = G (m_1+m_2) \frac{ ({\bf r'}_2 - {\bf r'}_1) }{ \left|{\bf r'}_2 - {\bf r'}_1\right|^3 }. $$ Multiplying both sides by $\mu=m_1m_2/(m_1+m_2)$, we recognize the classical equation for the relative motion $$ \mu {\bf a'}_{1} = G m_1m_2\frac{ ({\bf r'}_2 - {\bf r'}_1) }{ \left|{\bf r'}_2 - {\bf r'}_1\right|^3 } $$ where the reduced mass $\mu$ here appears as effect of the inertial force.