Cardinality of factors of infinite non-abelian groups

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 $G=A\cdot B_1=A\cdot B_2$.

Here's an example with $|B_1|=1, |B_2|=2$: Let $G=\mathbb Z*\mathbb Z$ (free product). Write non-identity elements of $G$ multiplicatively as reduced words with $X^{\mathbb Z-\{0\}}, Y^{\mathbb Z-\{0\}} $(with symbols $X,Y$). Let $A_0=X^{\mathbb Z - 2\mathbb Z}, A_1 = \{$reduced words with $X^{2\mathbb N},Y^{2\mathbb N}\}, A=A_0A_1, B_1=\{1\}, B_2=\{X^2,Y^2\}$. Note we can't have $A \cdot B_1=A \cdot B_2=G$ with $|B_1|=1, |B_2|>1$.