Why doesn't the Schrödinger equation contain a term of $mc^2$?

Just because to introduce the constant added term $mc^2I$ to the Hamiltonian operator would be equivalent to redefine $\psi \to \psi'= e^{imc^2 t/\hbar}\psi$. This sort of phases do not matter in QM. You cannot see them by measuring any observable. Pure states are actually operators of the form $|\psi \rangle \langle \psi|$ and you see that these phases cancel each other.

Instead, if the mass were replaced by a mass operator with discrete spectrum the picture would change. In the classical limit the rapid temporal oscillations of the phases (I am assuming that the mass is big if compared with the typical energies of the system), would destroy the coherence of superpositions of different masses giving dynamically rise to superselection rule of the mass Bargmann's superselection rule (see here or here).

In nonrelativistic quantum mechanics, particles cannot be created or destroyed, and each particle has constant mass $m$. That means the extra $E = mc^2$ energy is just a constant, so it can be subtracted out by adding a constant to the Hamiltonian; only energy differences matter.

The $E = mc^2$ can play a role in quantum field theory, since particles can be created or destroyed there; for example, it is released in pair annihilation, giving the products extra energy.

Let's say we start with this equation of yours :

$$-\frac {\hbar^2}{2m} \frac {\partial^2\psi}{\partial x^2}+V\psi +mc^2\psi=i\hbar \frac {\partial \psi}{\partial t}$$

Now a simple transform reduces it to the original form :

$$\psi = e^{Wt}\phi$$

$$-\frac {\hbar^2}{2m} e^{Wt} \frac {\partial^2\phi}{\partial x^2}+Ve^{Wt}\phi+mc^2e^{Wt}\phi=i\hbar e^{Wt} \frac {\partial \phi}{\partial t} + i\hbar W e^{Wt}\phi$$

Which reduces to :

$$-\frac {\hbar^2}{2m} \frac {\partial^2\phi}{\partial x^2}+V\phi+mc^2\phi=i\hbar \frac {\partial \phi}{\partial t} + i\hbar W \phi $$

And it is easy to see that $i\hbar W := mc^2$ recovers the original form of the equation.

So as we did in adding the rest mass term was add a rather pointless phase term that does nothing for us :

$$\psi = \phi \,\, exp\left( i\frac{mc^2}{\hbar} t\right)$$

This is the phase change that has no net effect that is discussed in the answers by Valter Moretti and Ruslan