To understand the "minimal troublemakers" in the case of solvable groups, I would start with a solvable group $G$ such that every proper section of $G$ has a complete prime graph but $G$ does not. Recall that a section of $G$ is a group $X/Y$ where $Y \lhd X$ and $X$ is subgroup of $G$.

Such a grooup $G$ must be a $\{p,q\}$-group for a pair of distinct primes $p$ and $q$ ( for otherwise $G$ has a Hall $\{p,q\}$-subgroup $H$ which is proper and $H$ contains an element of order $pq).$ Also, a Sylow $p$-subgroup of $G$ must be a maximal subgroup of $G$, and likewise a Sylow $q$-subgroup of $G$ must be maximal.

Now $G$ can't have both a normal Sylow $p$-subgroup and a normal Sylow $q$-subgroup. However, we have either $O_{p}(G) \neq 1$ or $O_{q}(G) \neq 1.$ Label so that $O_{p}(G) \neq 1.$ Then $G/O_{p}(G)$ must be a $q$-group ( otherwise it contains an element of order $pq$, and then so does $G$).

Hence $O_{p}(G)$ is a normal Sylow $p$-subgroup of $G$, and is hence a maximal subgroup of $G$, so that $[G:O_{p}(G)] = q.$ Hence $G$ is a Frobenius group with kernel $O_{p}(G)$ and cyclic complement of order $q$.

We can go a little further since a Sylow $q$-subgroup $Q$ of $G$ is a maximal subgroup. For then $G = QM$ for some minimal normal subgroup of $G$ contained in $O_{p}(G),$ and we deduce that $|G|$ has the form $qp^{e}$ where $e$ is the smallest positive integer such that $q$ divides $p^{e}-1.$

almostsimple. But, as you've mentioned, youcanhave $G$ being the direct product of 2 non-abelian simple groups, and that means that a full classification (using $F^*(G)$) is likely to be a bit tricky. $\endgroup$ – Nick Gill Sep 15 '16 at 8:59