Let $G$ be a finite group such that if $A, Ble G$ then $ABle G$. Prove $G$ is a solvable.

Here's another proof, now that you have the answer. Let $x$ and $y$ be elements of coprime order. Then $langle xranglelangle yrangle$ is a subgroup, and has order $o(x)cdot o(y)$. Thus any two elements of coprime order commute.

It turns out this is enough to prove the result: a finite group $G$ is nilpotent if and only if every $2$-generated subgroup is nilpotent.

But let's not use that, and let $z$ be an element in the centre of some Sylow $p$-subgroup of $G$. Consider $C_G(z)$. It contains all elements of order prime to $p$, and also contains a Sylow $p$-subgroup of $G$, thus is all of $G$. Thus $zin Z(G)neq 1$.

Since your condition is inherited by quotients, we see that $G$ must possess a central series, hence is nilpotent.

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Edit: some more on the structure of these groups. In general (i.e., infinite groups), they are locally nilpotent. The finite ones were mostly classified by Iwasawa in 1941, but there were some gaps that were filled by Napolitani and Janko.

A $p$-group is quasihamiltonian (i.e., $AB$ is a subgroup for all $A,Bleq G$) if and only if

1. It has an abelian normal subgroup $N$ such that $G/N$ is cyclic, and there exists a generator $x$ of $G/N$, and an integer $n$, such that $a^x=a^{1+p^n}$ for all $ain A$, or
2. $p=2$, and $G$ is the direct product of $Q_8$ and an elementary abelian $2$-group.

A subgroup $H$ such that $HK=KH$ (i.e., $HK$ is a subgroup) for all $Kleq G$ is called a permutable subgroup. The above groups are those where all subgroups are permutable. There's been recent research on groups where lots of, but not all, subgroups are permutable.

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Edit 2: here's another proof, which given you mention Burnside's $p^alpha q^beta$ theorem, might have been the intention of your tutor (assuming you are doing a course).

Let $xin G$ have order $p$ for some $pmid |G|$. By the above method, we see that $yin C_G(x)$ for all elements $yin G$ of order prime to $p$. But now this means that $|G:C_G(x)|$ is a power of $p$. By that useful theorem you did while proving Burnside's $p^alpha q^beta$ theorem, $G$ cannot be a non-abelian simple group. Thus $G$ possesses a normal subgroup $N$. The condition is inherited by both $N$ and $G/N$, so by induction both of these are soluble. Thus $G$ is soluble.

Claim: All Sylow subgroups of $G$ are normal.

Proof: Suppose $P_1,P_2$ are different Sylow $p$-subgroups of $G$. Then the group $P_1P_2$ is a $p$-group that is strictly larger than $P_1$, contradiction.

Corollary: $G$ is solvable.