Subgroup membership problem for $BS(1,2)$ is decidable, as answered in this link: https://math.stackexchange.com/a/3220976/465070
$BS(1,2)$ is isomorphic to a subgroup of $GL(2,\mathbb{Q})$ (the group of $2 \times 2$ rational matrices with determinant $\pm 1$).
Are there any other known results about the subgroup membership problem (or semigroup membership problem) for other rational matrix groups?
The subgroup membership problem is not decidable for groups which are linear over $\mathbb{Z}$ (that is, group of integer matrices).
The Mihailova subgroup of direct products. Note that $F_2\times F_2$ is linear over $\mathbb{Z}$, where $F_2$ is the free group of rank two. A striking example of a finitely generated subgroup with undecidable membership problem is the "Mihailova subgroup" of $F_2\times F_2$ (the reference is K. A. Mihailova, The occurrence problem for direct products of groups Dokl. Acad. Nauk SSRR 119 (1958), 1103-1105.). The idea is as follows: take a surjective homomorphism $\phi:F_2\rightarrow G$ where $G=\langle \mathbf{x}\mid\mathbf{r}\rangle$ has insoluble word problem and consider the diagonal subgroup $$\Delta=\{(g, g)\in G\times G\mid g\in G\}.$$ Then the subgroup membership problem for $\Delta$ is not decidable in $G\times G$, and hence the membership problem for $Q=(\phi\times\phi)^{-1}(\Delta)$ is not decidable in $F_2\times F_2$. Moreover, $Q$ is finitely generated; it is generated by the set $$\{(x, x)\mid x\in\mathbf{x}\}\cup \{(R, 1)\mid R\in\mathbf{r}\}\cup \{(1, R)\mid R\in\mathbf{r}\}.$$
Rips' construction. A different kind of example can be found using Rips' construction. I wrote a length post about this construction here, which I don't want to repeat. The idea is that for every finitely presentable group $Q$ there exists a "small cancellation" group $H$ and a two-generated subgroup $N$ such that $H/N\cong Q$. So if $Q$ has insoluble word problem then the membership problem for $N$ is undecidable. The issue here is that it is not obvious that $H$ is linear over $\mathbb{Z}$; this was a major result of Dani Wise and his co-authors (although some years before he made his fame and fortune, before he proved that all small cancellation groups are linear over $\mathbb{Z}$ and a million other amazing things, Wise wrote a paper called "a residually finite version of Rips' construction" which is all you need here :-) ).