Let $G=C_2\ast C_3$ be the free product of cyclic 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 left coset representatives for both $B_1=\langle a\rangle$ and $B_2=\langle b\rangle$, so $G=A\cdot B_1=A\cdot B_2$.