Does there exist a field $k$ and a subring $R$ of $S = M_2(k)$ such that $R$ is not finitely generated over its center, $S=kR$ and $1_R = 1_S$? ($S$ is the algebra of $2 \times 2$ matrices over $k$.)
I think the answer is "yes". Let $A$ be a nonNoetherian integral domain (for example a polynomial ring in infinitely many variables over a field), let $I$ denote a nonfinitelygenerated ideal, and let $k$ be the field of fractions of $A$. Let $R$ denote the ring of $2\times 2$ matrices with coefficients in $A$ and with bottom left hand entry in $I$. I think this ticks all the boxes. For example $kR=M_2(k)$ because I can scale any element of $M_2(k)$ until it's in $M_2(R)$ and then again so that all entries are in $I$. However, I don't think $R$ can be finitelygenerated over its centre (which is easily checked to be $A$). For if $r_1,r_2,\ldots,r_n$ are finitely many elements of $R$ then the ring they generate over $A$ will be contained in the $2\times 2$ matrices with coefficients in $A$ and bottom left hand entry in $J$, the finitelygenerated ideal generated by the bottom left hand entries of the $r_i$, and this is a proper subset of $I$. 


This question has been explored in the context of polynomial identity rings (PIrings). Your hypothesis implies that $R$ is a prime PIring of PIdegree 2 (see below). However, the main structure theorems of PItheory, Kaplansky's Theorem, Posner's Theorem, ArtinProcesi Theorem and central polynomials, are in the opposite direction. They show that if a ring $R$ satisfies a polynomial identity plus a suitable further hypothesis, then $R$ is, or almost is, a finite module over its center, or at least has a large center. One example is the strong form of Posner's theorem using central polynomials: Let $R$ be a prime PIring with center $C$, and let $T = C  \{0\}$. Then for some integer $n$, $T^{1}R$ is a central simple algebra of finite dimension $n^2$ over its center $T^{1}C$. The theory of central simple algebras says that if $A$ is finite dimensional central simple over a field $L$, then its dimension over $L$ is a square $n^2$, and there is a unit preserving $L$algebra embedding into $M_n(k)$ for some extension field $k$ of $L$. Combining this theory with Posner's theorem gives: A ring $R$ (with unit) is a prime PIring if and only if there is a field $k$, an integer $n$, and a unit preserving ring embedding $R \to M_n(k)$ such that $kR = M_n(k)$, where $R$ is identified with its image in $M_n(k)$. The field $k$ is not unique, but the integer $n$ is. It is called the PIdegree of $R$. Thus your hypothesis: "$R$ is a subring of $S = M_2(k)$, where $k$ is a field, $S = kR$, and $1_R = 1_S$", implies that $R$ is a prime PIring of PIdegree 2. One result showing that a prime PIring is close to being a finite module over its center is a theorem of mine (p. 174 in DrenskyFormanek, "Polynomial Identity Rings"): If $R$ is a prime PIring of PIdegree $n$ with center $C$, then there is a $C$module embedding of $R$ into a free $C$module of rank $n^2$. As for your question, there is an example due to Cauchon (p. 228 in Rowen, "Polynomial Identities in Ring Theory") of a Noetherian prime PIring of PIdegree 2 which is not a finite module over its center. Cauchon also proved that if a prime PIring has the ascending chain condition on twosided ideals, then it has the ascending chain condition on left and right ideals. In other words, left Noetherian, right Noetherian, and ACC on twosided ideals are equivalent for prime PIrings. 

