This is a continuation of Are these exactly the abelian groups? I would like to consider another condition on a group and see if it implies commutativity. The condition is
$$(\forall A,B\subseteq G)(\forall g\in G)(\exists x,y\in G)\ AgB=xABy.$$
Again, abelian groups satisfy this condition -- it's enough to take $x=g$ and $y=1.$
Using an argument similar to Hurkyl's in the other question, I've been able to prove that if two elements $g,h$ of such a group $G$ do not commute then they must generate the quaternion group. I've also checked that the quaternion group does not satisfy the condition, but this is not enough: the condition could still hold in $G$ even if it doesn't hold in $\langle g,h\rangle.$
I haven't been able to do anything more. Considering other equations just left me with loads of unmanageable cases in which I simply got lost.
If, in the quaternion group, you can find an example where $$\left\vert AgB\right\vert\neq\left\vert AB\right\vert,$$ then $A,B,g$ will contradict the condition in any group containing the quaternion group.
This is probably not the smallest example, but I found $A=B=\{i,j,k\}$, $g=i$, where $$AB=\{-1,i,-i,j,-j,k,-k\}$$ has seven elements, but $$AgB=\{1,-1,i,-i,-j,-k\}$$ has six elements.