Here's what I know about Question 1 from tinkering with the linked question for awhile.

- If $R$ is an algebraically closed field of characteristic zero then the classification is the same as the classification of the finite subgroups of $\text{SL}_2(\mathbb{C})$: the cyclic groups, the dicyclic groups, and the binary polyhedral groups.  Hence if $R$ is an integral domain of characteristic zero then any finite subgroup must be on this list.

- If $R$ is an integral domain of characteristic $2$ then $\text{SL}_2(R)$ has no nontrivial elements of order $2$.  This rules out any finite group of even order.

- If $2$ is not a zero divisor in $R$ (in particular, if $R$ is an integral domain) then any element of order $2$ in $\text{SL}_2(R)$ is a scalar multiple of the identity, in particular central.  This rules out many finite groups, including non-abelian simple groups by Feit-Thompson.

- If $2 = 0$ in $R$, then an argument in my answer to the linked question shows that $\text{SL}_2(R)$ cannot contain a non-abelian simple group with an element of order $4$. 

- In general, let $I$ be the ideal of $R$ consisting of the elements annihilated by $2$.  The above bullet point shows that any element of order $2$ in the quotient $\text{SL}_2(R/I)$ must be a scalar multiple of the identity, in particular central, in this quotient.  Hence if $G$ is a non-abelian simple subgroup of $\text{SL}_2(R)$, it must land in the kernel of the above map: the subgroup congruent to the identity $\bmod I$.  

I strongly suspect that $\text{SL}_2(R)$ can't contain a non-abelian simple group in general but have not yet been able to prove it.  It just seems as if there is not enough room to be "too noncommutative."