A complete answer seems not to be known. Let me give you the following two nearly-contemporaneous references from the mid-70s:
Robert Guralnick, On groups with decomposable commutator subgroups
Michael Miller, Existence of Finite Groups with Classical Commutator Subgroup
Both Guralnick and Miller call groups which are commutator subgroups $C$-groups (though I don't know who, if either, originated the term) and give partial answers to your general question. For example, Theorem 4 from Miller gives the following:
Let $G$ be a subgroup of $\operatorname{GL}_n(K)$ containing $\operatorname{SL}_n(K)$ for $K$ a finite field of characteristic not equal to 2. Then $G$ is the commutator subgroup of some group unless it is of odd index and $n$ is even.
The groupprops-wiki calls such groups commutator-realizable, and give a basic result on such groups, but mention that this terminology is not standard (though is probably safer than the overloaded term $C$-group.)
Edit: Some googling around led to the following slick argument of Schoof (from his Semistable abelian varieties with good reduction outside 15), which is closely related to your observation in bullet (3), and also serves to eliminate the symmetric groups. I'll quote verbatim except for change of variable names:
Let $G$ be a group and let $G'$ be its commutator subgroup. Conjugation gives rise to a homomorphism $G \to \operatorname{Aut}(G')$. On the one hand it maps $G'$ to the commutator subgroup of $\operatorname{Aut}(G')$. On the other hand the image of $G'$ is the group $\operatorname{Inn}(G')$ of inner automorphisms of $G'$. Therefore, if a group $X$ is the commutator subgroup of some group, we must have $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$.
