Let $K=\mathbb Z_2 \times \mathbb Z_2$. What finite solvable groups (up to isomorphism) have Fitting subgroup isomorphic to $K$?
So far I've found that $K, S_4, A_4$ satisfy that property. I don't know whether there are others... What follows is what I know. Let $G$ be a group we want to find. Then there is $n\in \mathbb N$ such that $G^{(n)}={1_G}$ (a derived subgroup is trivial). Its Fitting subgroup is the intersection of all the $p$-cores of $G$, or equivalently, the unique maximal normal nilpotent subgroup.
Since $p$-groups are nilpotent, if $G$ is a $p$-group, then $G=K$. Let's assume $G$ is not a $2$-group. Then there is $q\neq 2$ prime, dividing $|G|$. For every $p\neq 2$ dividing $|G|$, there are more than one Sylow $p$-subgroups, since its core is trivial.
Another important fact about finite solvable groups is that the centralizer of the Fitting subgroup is its center. In our case $C_G(K)=K$. Yet I can't see how this would help.
Is there a general strategy for any group $K$?
Since $C_G(K)=K$, the conjugation action of $G/K$ on $K$ maps $G/K$ into a subgroup of ${\rm Aut}(K) \cong S_3$. We can't have $|G/K| = 2$, so $|G/K| = 1,3,$ or $6$ giving the three examples that you have found. So there are no more examples.
The same strategy would work for any abelian $p$-group $K$. Otherwise it would get more complicated but could still be done because $G/Z(K)$ must be a subgroup of ${\rm Aut}(K)$, and hence $|G|$ is bounded.