Does a finite product of commutators of the form $[a,b]$ lie in $A∩B$?

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1) If $G = A★B$ be the free product of two groups $A$ and $B$, then $A$ and $B$ need not be normal to $A★B$ necessarily, right?

My Aim:

I need to know if finite product of commutators of the form $[a,b]$ with $a ∈ A$ and $b ∈ B$ lie in $A ∩ B$ even without above normality.

My larger Aim:

Consider the free group $\langle G,G\rangle$ freely generated by all pairs $\langle x,y\rangle$ with $x,y ∈ G$. I am trying to prove that given $M =$ the sub group generated by $\{ \langle a,b\rangle \mid a ∈ A/\{1\}, b ∈ B/\{1\} \}$ and $μ ∈ M$ such that $[μ] = 1$, prove that $μ = 1$.

I can prove the above if $A$ and $B$ are normal sub groups of $G$, but unlucky otherwise. Any help?