Does such a group exist? A group with infinite normal subgroups:
$G\varsupsetneqq G_{E}\varsupsetneqq G_{1}\varsupsetneqq G_{2}\varsupsetneqq....$
such that there exist a subset $B\subset G$ satisfying:
1) $\forall x,y\in B\,\, x\neq y\Rightarrow x\cdot y^{-1}\notin G_{E}$
2) $\forall n\in\mathbb{N}$  $G/G_{n}=\{{\overline{x}\cdot\overline{y}\cdot\overline{z}\,|\, x,y,z\in B\}}$
 A: $\mathbb Z \times \mathbb Z$ will do. Take $G_E$ to consist of the elements whose first coordinate is $0$ and $G_n$ to be the subgroup of multiples of $2^n$ inside $G_E$. What $G_n$ are is immaterial, since we will make $G$ itself be covered by $\{x\cdot y\cdot z| x,y,z\in B\}$.
We will do this by building up $B$ step-by-step. At any given time, it contains finitely many elements. Choose some element of $G$ that is not in $\{x\cdot y\cdot z| x,y,z\in B\}$. We will add $3$ elements to $B$ so that it is. We can do this while preserving the property (1), since that just means that the first coordinates of all the elements are different. So choose two very large positive first coordinates, and then a third very large negative first coordinate so that they add up to the desired value. If they are sufficiently large they will not intersect past first coordinates. Then choose second coordinates in any way that adds up to the desired value.
Repeat this process until every element of $G$ is covered. Then clearly $G/G_n$ is covered as well.
This should work any time $G/G_E$ is infinite (of at least the cardinality of $G_E$).
