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:

<blockquote>
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.
</blockquote>

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:

<blockquote>
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 its image 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 $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$.
</blockquote>