It occurs to me that $R^\times$ (the group of units of a commutative ring) may have a subgroup, say $S \leqslant R^\times$.
It seems that we could then define the group $$ \operatorname{GL}_n(R; S) = \{A \in\operatorname{GL}_n(R) : \det A \in S\} $$ Does this group make sense? Is something like this studied? Is there anything interesting about $GL_n(R; S)$? The only thing I can think of is that we of course have: $$ \operatorname{SL}_n(R) \leqslant\operatorname{GL}_n(R; S) \leqslant \operatorname{GL}_n(R) $$
Some notes
As discussed in the comments of the accepted answer, the subgroups of $\operatorname{GL}_n(R)$ containing $\operatorname{SL}_n(R)$ are precisely the $\operatorname{GL}_n(R; S)$.
For all $n \geq 2$, there are subgroups of $\operatorname{GL}_n(R)$ which neither contain nor are contained by $\operatorname{SL}_n(R)$. An example is $$ \left\{\begin{pmatrix} a & 0 \\0 & \mathbb{I}_{n-1} \end{pmatrix} : a \in R^\times \right\} \leqslant\operatorname{GL}_n(R) $$
Lemma: If $f:\ A\ \longrightarrow\ B$ is a group homomorphism, then for every subgroup $B'\subseteq B$ the subset $A':=f^{-1}(B')$ is a subgroup of $A'$ containing $\ker f$. (This is an easy exercise in group theory.)
In particular, the determinant map on $\operatorname{GL}_n(R)$ is a group homomorphism $$\det:\ \operatorname{GL}_n(R)\ \longrightarrow\ R^{\times},$$ with kernel $\operatorname{SL}_n(R)$. So every subgroup of $S\subset R^{\times}$ defines a subgroup of $\operatorname{GL}_n(R)$ containing $\operatorname{SL}_n(R)$ by taking the inverse image under the determinant map: $$\det{}^{-1}(S)=\{A\in\operatorname{GL}_n(R):\ \det A\in S\}.$$ These are indeed precisely the subgroups you describe. There may of course be more subgroups.