Let $S \subset R$ be two non-commutative rings and assume that $R$ is free of finite rank as an $S$-module and that $S$ is central subring of $R$. What are the (minimal) conditions such that:
$${\rm Hom}_S(R,S) \cong R\text{ as $R$-modules} ? $$
Here $R$ acts on ${\rm Hom}_S(R,S)$ via $r. \phi(x)=\phi(xr)$. One may prove that ${\rm Hom}_S(R,S) \cong R$ as $S$-modules.
For example, if $S=k$ is a field and $R$ is a finite-dimensional $k$-algebra, the condition is equivalent to $R$ being a Frobenius algebra. Any references are welcome.
Best Answer
Not sure if this is what you are looking for, but certainly this is a necessary condition for your condition to hold.
Let $S$ be a central subring of a (not necessarily commutative) ring $R$, with $R$ free and finite dimensional as a module over $S$. Your condition is that $R$ is isomorphic to ${\rm Hom}_S(R,S)$ as left $R$ modules. Equivalently, there exists a map $\epsilon\colon R\to S$ such that every $S$-linear homomorphism $R \to S$ may be written in the form $\epsilon(\_a)$ for a unique $a\in R$.
A necessary condition for such an $\epsilon$ to exist is that finitely generated projective $R$-modules are injective relative to $S$. That is given an $R$-linear map of left $R$ modules $f\colon A \to M$ such that $f$ has a left inverse as a map of $S$ modules, any $R$-linear map $h\colon A \to P$ (for $P$ a finitely generated projective module) may be extended to an $R$-linear map $M \to P$.
$$ A\stackrel f\to M $$ $$h\downarrow \,\,\,\swarrow\quad$$ $$P\quad\quad$$
Proof: Suppose $\epsilon$ exists as above. It is sufficient to consider the case $P=R$, as the property of being relatively injective extends in an obvious way to (finite) direct sums and summands.
Given $m\in M$ we have an element of ${\rm Hom}_S(R,S)$ given by $$\lambda\mapsto \epsilon(hg(\lambda m))$$ where $g$ is the $S$-linear left inverse to $f$.
Thus we have $\hat h(m)\in R$ such that $$\epsilon(hg(\lambda m))=\epsilon(\lambda \hat h(m),$$ for all $\lambda\in R$. Then $\hat h$ is $R$-linear as for all $\lambda\in R$ we have $$\epsilon (\lambda \hat h(\mu m))=\epsilon(hg(\lambda\mu m))=\epsilon(\lambda\mu\hat h(m)).$$
Finally we note that $\hat hf=h$: $$\epsilon(\lambda\hat hf(a))=\epsilon(hg(\lambda f(a)))=\epsilon(hgf(\lambda a))=\epsilon(h(\lambda a))=\epsilon(\lambda h(a)),$$ for all $\lambda \in R, \,\, a \in A$.