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 small, complete category you get by taking sheaves over, say, $[0,1]$. 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?

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 any small, complete (that's small AND complete) category 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.

How does that look?

Ok, since no one has answered this, I'd like to at least give an attempt. Here is my intuition for why I think it's false (I say intuition since I haven't verified that everything here works).

Okay: If $V: A \rightarrow R-Mod$ preserved arbitrary products then, since it is exact, it would (I think?) preserve small limits. Now, (and this is the big part I haven't checked) if either the adjoint functor theorem or special adjoint functor theorem was applicable here, we would get a left adjoint $R-Mod \rightarrow A$. Left adjoints of exact functors preserve projective objects, and this just doesn't seem kosher if you want a generalization of Mitchell's embedding for every single abelian category $A$. In particular it doesn't seem kosher for the category of sheaves over a reasonable space (locally connected Hausdorff, maybe?), since the only projective objects in such a category are 0, and I'm sure we can't have all the projectives mapping to zero in this left adjoint or something would go wrong.

Sorry that this isn't fully worked out, but it was too long to leave as a comment... Some help? Suggestions?

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 small, complete category you get by taking sheaves over, say, $[0,1]$. 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?