Regarding the original question (with $A=R$-$\mathbf{Mod}$), I think that
by SAFT *any* continuous functor $A^{\mathrm{op}}\to \mathbf{Set}$ is representable,
and hence the assertion in the original question does not generalize
Baer's theorem.

In detail (with $A=R$-$\mathbf{Mod}$):

(*) $R$ is a generator in $A$, and hence a cogenerator in
$A^{\mathrm{op}}$.

(*) $A$ is co-well-powered, because there is a bijection between the
quotient objects of $M\in A$ and the set of submodules of $M$, and the
latter set is small (since by assumption $M$ is small). It follows
that $A^{\mathrm{op}}$ is well-powered.

(*) $A$ is small cocomplete (as is any $\tau$-algebra, for
$\tau=$(operations, identities)), and hence $A^{\mathrm{op}}$ is small
complete.

(*) Both $A^{\mathrm{op}}$ and $\mathbf{Set}$ have small hom-sets.

So, all the conditions of SAFT hold for a functor
$A^{\mathrm{op}}\to\mathbf{Set}$,
and hence any such continuous functor has a left adjoint. Now, if a
functor $G\colon A^{\mathrm{op}}\to\mathbf{Set}$ has a left adjoint
then it is surely representable: Saying that a functor $G$ is
representable is like saying that there is a universal arrow from a
one-object set $1$ to $G$ (Prop. 3.2.2, p. 60 in Mac Lane), and for
this we can take the unit $\eta_1\colon 1\to G(F1)$ (with $F$ the left
adjoint of $G$).

(See also the discussion on Watt's theorem on p. 131 of Mac Lane).

I am not sure about the general case of the edited question
(where $A$ is an arbitrary abelian category with a generator +
AB3--AB5). Cocompleteness holds by AB3 (as I have seen in
Wikipedia
), but I do
not know enough to say anything about the question of being co-well-powered.