About the product of finite subsets of a torsion free group

Question

Let $G$ be a torsion free group with identity $e$. For a subset $X$ of $G$, denote by $X^\#$ the set $X\setminus\{e\}$. Let $A$ be a finite subset of $G$ containing $e$. Is there a finite subset $B$ containing $e$ such that $$A\subset B^\#A\quad\text{and}\quad B\subset BA^\#$$ It is obvious there is no such $B$ when $A=\{e,x\}$; this follows from the right inclusion and that $G$ is torsion free, so the $A$ in my question has at least 3 elements.

The above question comes from the zero divisor conjecture; if there exist non zero $\alpha,\beta\in\mathbb C[G]$ such that $\alpha\beta=0$, without loss of generality we can assume $A:=\text{support}(\alpha)$ and $B:=\text{support}(\beta)$ contain $e$, and then $A$ and $B$ should satisfy in above inclusions.

Answer 1

If such $B$ exists then for some $n>0$ there must exist elements $a_1,\dots ,a_n\in A^\#$ such that $a_1\dots a_n=e$. (The argument is essentially the one you use to rule out the case $\lbrace e,x\rbrace$ when $G$ is torsionfree.)

Conversely, suppose that there exist $a_1,\dots ,a_n\in A^\#$ such that $a_1\dots a_n=e$. Then $B$ can be taken to be $A^n=A\dots A$. In fact,

(1) for any $a\in A$ we have $a=ae\in B^\#A$ if $a\neq e$ and $a=(a_1\dots a_{n-1})a_n\in B^\#A$ if $a=e$,

(2) any $b\in B$ (including $e$) is a product $x_1\dots x_k$ with all $x_j\in A^\#$ and $1\le k\le n$, so that $b=(x_1\dots x_{k-1})x_k\in BA^\#$.