This is an extension of this MSE question, in which I asked whether there was a counterexample to the following statement,

Conjecture.If a finite group $G$ contains a $\lbrace p,q \rbrace$-Hall subgroup for every pair of primes $p$ and $q$ dividing $|G|$, then $G$ is solvable.

which is a refinement of the converse to Hall's theorem,

Theorem (Hall).Denote by $\pi(G)$ the set of prime divisors of $|G|$. Then $G$ is solvable if and only if $G$ contains a $\pi$-Hall subgroup for every subset $\pi$ of the prime divisors of $|G|$.

*Edit:* As requested, we call $H\leqslant G$ a $\pi$-Hall subgroup if $|H|$ and $[G:H]$ are coprime and $p$ divides $|H|$ for every $p\in \pi$. So, Hall subgroups are a generalization of Sylow subgroups for multiple primes.

I received a great answer from Geoff Robinson, who thinks there probably isn't counterexample, and proposed to check it case-by-case using the classification of finite simple groups. I am still digesting his answer, however this led me to wonder whether (assuming the statement is true) there is a proof that does not rely on the classification theorem.

My original idea, before I posted the question on MSE, was to was to show that this implies the hypothesis to the to the original converse - that is, $G$ contains a $\pi$-Hall subgroup for every $\pi\subseteq \pi(G)$ iff it contains a $\{ p,q \}$-Hall subgroup for each $p,q\in \pi(G)$ - but I haven't found any way to make this work yet. It has also been suggested to me to try to mimic the original inductive proof, which (for $|\pi(G)|\geq 3$) makes use of this lemma, but again I do not see how to put it together.

So, my question is, is this conjecture provable without using the classification of finite simple groups?

If it is false, is there a counterexample of a non-solvable group with $\lbrace p,q\rbrace$-Hall subgroups for every pair of primes (which is thus missing some other Hall subgroup, e.g. a $\lbrace p,q,r \rbrace$-Hall subgroup)?