The equivalent proposition is:
Proposition. $R$ and $S$ are Morita equivalent if and only if there is a progenerator $P$ of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$.
While searching for the answer of this question, I found some texts in the internet which stated the above proposition putting $\text{End}(P) \simeq S$ instead of $\text{End}(P) \simeq S^{\text{op}}$, and it is wrong! I thought it was important to share this detail, because it may confuse people, as it confused me. Now let's see the explanation:
The fundamental motive for the appearence of $S^{\text{op}}$ instead of $S$ is the way we compose functions. We have the following (which is easily verified):
If we regard the ring $S$ as a right module ($S_{S}$), then $\text{End}(S_{S}) \simeq S$ and if we regard $S$ as a left module ($_{S}S$), then $\text{End}(_{S}S) \simeq S^{\text{op}}$ (those are ring isomorphisms).
Now, for more details, let us examine the necessity of those propositions closely:
If $F : \text{Mod}_{S} \rightarrow \text{Mod}_{R}$ is an equivalence of categories, then putting $P_{R} = F(S_{S})$, we may show that $P_{R}$ is a progenerator, and to verify that $\text{End}(P_{R}) \simeq S$, we proceed like this: $$\text{End}(P_{R}) = \text{End}(F(S_{S})) \simeq \text{End}(S_{S}) \simeq S.$$
And for the left case we have:
If $F : \text{ }_{S}\text{Mod} \rightarrow \text{ }_{R}\text{Mod}$ is an equivalence of categories, then putting $_{R}P = F(_{S}S)$, we may show that $_{R}P$ is a progenerator, and we have: $$\text{End}(_{R}P) = \text{End}(F(_{S}S)) \simeq \text{End}(_{S}S) \simeq S^{\text{op}}.$$
To prove the sufficiency of the propositions, we proceed as follows:
If $P$ is a progenerator of $\text{Mod}_{R}$ such that $\text{End}(P) \simeq S$, then we may show that $$\text{Hom}_{R}(P,-) : \text{Mod}_{R} \rightarrow \text{Mod}_{S}$$ is an equivalence of categories, and for $M$ in $\text{Mod}_{R}$, we regard $\text{Hom}_{R}(P,M)$ as a right $\text{End}(P)$-module in the usual way, by composing functions, so that it becomes a right $S$-module.
For the left case we have:
If $P$ is a progenerator of $_{R}\text{Mod}$ such that $\text{End}(P) \simeq S^{\text{op}}$, then we may show that $$\text{Hom}_{R}(P,-) : \text{ }_{R}\text{Mod} \rightarrow \text{ }_{S}\text{Mod}$$ is an equivalence of categories, and for $M$ in $_{R}\text{Mod}$, we regard $\text{Hom}_{R}(P,M)$ as a left $\text{End}(P)^{\text{op}}$-module in the usual way, by composing functions and making the necessary adjustments (putting the $^{\text{op}}$) so that everything works, and then it becomes a left $S$-module, since $\text{End}(P)^{\text{op}} \simeq (S^{\text{op}})^{\text{op}} = S$.
See this question and answer for a proof. The idea is exactly the one described in the second part of Andrea's answer: for any $A$-module, you have a presentation $A^{(J)}\to A^{(I)}\to M\to 0$, which you can then use together with your assumptions to deduce that $F(M)\cong F(A)\otimes_A M$.
Note that $F(A)$ has a $B$-module structure by definition, and also a right $A$-module structure given by $F(r_a), a\in A$ where $r_a : A\to A$ is given by $x\mapsto xa$ (this is a morphism of left $A$-modules).
To get naturality of the isomorphism $F(M)\cong F(A)\otimes_A M$, the trick is to define, for any $F$ and any $M$ (no assumptions so far) a natural morphism $F(A)\otimes_A M\to F(M)$. One then uses the assumptions to prove that this natural morphism is an isomorphism.
This natural morphism is defined by the following composite: $M\to \hom(A,M) \to \hom(F(A),F(M))$ and then the universal property of the tensor product. There is one thing to note here : $\hom(A,M)\to \hom(F(A),F(M))$ is a morphism of $A$-modules because of the definition of the right $A$-module structure on $F(A)$.
This gives you a natural morphism, which is obviously an isomorphism when $M=A$, and since both sides preserve direct sums and cokernels, it is an isomorphism for any $M$.
Let me mention two additional things with respect to some points in Andrea's answer :
1- In the example he gives in the first part of the question, there is a mistake : we never consider $P\otimes_\mathbb C X$, but $P\otimes_A X$ - in particular the dimension is not the product of the dimensions.
2- There is no difference between preserving cokernels of general maps and preserving cokernels of injections: indeed if you preserve the latter, then you preserve epimorphisms, and then for a general cokernel sequence $M\to N\to C\to 0$, you can separate it as $M\to \mathrm{im}(M\to N)\to 0$ and $0\to \mathrm{im}(M\to N)\to N\to C\to 0$.
This gives two a priori different definitions of right exact functors which end up agreeing.
Best Answer
Yes. A simple module is an object $A$ of $\mathrm{Mod}_R$ which is not a zero object such that every monic morphism $B\to A$ from any other object $B$ either factors through a zero object or is an isomorphism. (In concrete terms, this says any injective homomorphism to $A$ is either $0$ or surjective, so every submodule of $A$ is either trivial or all of $A$.) Each part of this definition is preserved by an equivalence of categories, so an equivalence of categories $\mathrm{Mod}_R\to \mathrm{Mod}_S$ sends simple modules to simple modules.