If $AB$ contains a neighborhood of $x$, does $m(A \cap x B^{-1})>0$?

This is a converse to a step in Steinhaus theorem for topological groups, where it is shown that $m(A \cap xB^{-1})>0$ implies $AB$ contains a neighborhood of $x$.

Suppose $m(A),m(B)>0$ on some locally compact topological group with left Haar measure $m$. Suppose $AB$ contains a neighborhood of $x$. (By Steinhaus theorem, there is always such an $x$). Can one conclude that $m(A\cap xB^{-1})>0$?

Consider the case of $\mathbb R$. Let $A = B = E \cup [10,11]$ where $E$ is the Cantor set. As is well known, $E + E = [0,2]$ so we can take $x = 1$. But $A \cap (1 - B) = E$ has measure $0$.