EDIT: I'm completely rewriting this in detail now that I think I've worked it out. Lots of possibilities for mistakes here, so stay alert!
I claim that the generalization to include $(a)$ fails. My counterexample goes like this... Take $A$ to be the any small, complete (that's small AND complete) category you get by taking sheaves over, say, $[0,1]$. with no nonzero projectives. If the embedding into $R-mod$ given by Mitchell preserved arbitrary products, then it would be continuous since $A$ has equalizers and any limits can be built from products and equialisers (where equalisers are preserved by exactness).
Now, for each $x \in R-mod$, consider the index set $I = {f: x \rightarrow Va \vert a\in A} = \bigcup_{a \in A} Hom(x, Va)}$. (This is a set since hom-sets are small, and all of $A$ is a set.) Now, any $x \rightarrow Va$ can be realized as $x \rightarrow Va \rightarrow Va$ (with the identity), so we have verified the "solution set condition" of the adjoint functor theorem. If all is good, we may conclude that this embedding has a left adjoint $R-mod \rightarrow A$.
Now, left adjoints of exact functors preserve projective objects (see Weibel). Now, if we choose some $b \in A$ that doesn't map to zero under the embedding $V$, and some free module $a \in R-mod$ that maps nontrivially to $b$, then by the bijection on hom-sets we get from the adjunction, we may conclude that $a$ maps to some nonzero element in $A$. But this is a contradiction, as the only projective elements of $A$ are zero(see Bredon's Sheaf Theory, for example).
How does that look?
