Reducible finite subgroups $G$ of ${\rm SL}(2,\mathbb{C})$ are Abelian, so these can be dealt with easily. Irreducible, but imprimitive, subgroups of ${\rm SL}(2,\mathbb{C})$ have an Abelian normal $A$ subgroup of index $2$. By unimodularity, $A$ is even cyclic. Furthermore, (again by unimodularity), $A$ contains every element of order $2$ in $G$, so $A$ is not complemented. If $A$ is not a $2$-group, then $G$ has a (cyclic) normal subgroup $H$ of odd order which is complemented by a Sylow $2$-subgroup $S$, so $G$ is a semidirect product $HS$.

If $A$ is a $2$-group, then $G$ has only one element of order $2$, and $G$ is a generalized quaternion group, so is not  a semidirect product.

This leaves irreducible primitive groups. Such a group $G$ has all Abelian normal subgroups central, so of order $2$ by unimodularity and Schur's Lemma. If $G$ is solvable, then $F(G)$ is a $2$-group, (but $G$ is not a $2$-group by primitivity), so $F(G)$ is quaternion of order $8$ ( any larger generalized quaternion $2$-group has automorphism group  a $2$-groups, so can't be $F(G)$ as $G$ is not a $2$-group). Then unimodularity forces either $G \cong {\rm SL}(2,3)$ or $G$ a binary octahedral group ( with generalized quaternion Sylow $2$-subgroup). The first possibility is a semidirect product, the second is not.
Unimodularity excludes $G \cong {\rm GL}(2,3)$.

If $G$ is not solvable, then a minimal normal subgroup of $G/Z(G)$ must be simple. I won't go through all arguments, but the only possibility that occurs is $G \cong {\rm SL}(2,5)$, as is well known, and this is not a semidirect product. Note that unimodularity (among other things) excludes $G \cong {\rm GL}(2,5).$