In a group, a finite left translation of $B$ covers $A$. Does any finite right translation of $B$ cover $A$?

Question

Let $G$ be a group and $A,B\subseteq G$. Suppose there's some finite set $F\subseteq G$ such that: $$A\subseteq FB$$

Is there any finite set $F'\subseteq G$ such that $$A\subseteq BF'$$ ?

Answer 1

No. Let $G:=\mathcal F(x,y)$ be the free group on $2$ generators, and let $$\begin{align} A &:=\{xy^n\,|\,n\in\Bbb N\} \\ B &:=\{y^n\,|\,n\in\Bbb N\}\,. \end{align} $$ Then, with $F=\{x\}$ we have $A\subseteq FB$, but if $A\subseteq BF'$, then for all $xy^n$ there is an $m$ such that $y^{-m}xy^n\in F'$. And, whatever $m$'s may be, these are all distinct because of $n$ and freeness.