Let $G=C_2\ast C_3$ be the free product of cyclccyclic groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$.
Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced words ending with $a$ except for $$ba,baba,bababa,\dots.$$
Then $A$ is a set of rightleft coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.