Let $G$ be a "nice" infinite group: at least finitely presented and residually finite, maybe also linear and right-orderable (or even bi-orderable, or residually free nilpotent).
Consider an element $\lambda$ in the group ring $\mathbb Z[G]$ which is "residually invertible", ie every image $\overline\lambda\in\mathbb Z[G/H]$, $H$ a normal subgroup of finite index, is invertible. Is $\lambda$ itself invertible ?
Motivation: one can generalize the question to matrices in ${\rm M}_n(\mathbb Z[G])$, and also replace $\mathbb Z[G]$ by its Novikov completion in the direction of a nonzero morphism $u:G\to\mathbb R$ :
$$\mathbb Z[G]_u=\{\sum_{n=0}^\infty a_ng_n: a_n\in\mathbb Z,g_n\in G,u(g_n)\to+\infty\}.$$
A positive answer to the analogous question on detecting invertible matrices in ${\rm M}_n(\mathbb Z[G])_u$, for $G=\pi_1(M)$ with $M$ a closed $3$-manifold, would have the following consequence: a nonzero class $u\in H^1(M,\mathbb Z)={\rm Hom}(\pi_1(M),\mathbb Z)$ would be represented by a fibration $M\to S^1$ if and only if every twisted Alexander polynomial associated to a finite covering of $M$ is unitary (= bi-unitary).
Remarks. I have a proof for $\mathbb Z$ (!), elementary but not completely obvious. This implies the result (also for matrices) if $G$ is virtually free Abelian. From this one can prove the result for $G={\rm Heis}_3(\mathbb Z)$, the $3$-dimensional Heisenberg group over $\mathbb Z$ (matrices $\pmatrix{1&x&z\cr0&1&y\cr0&0&1}$ with $x,y,z\in\mathbb Z$), the simplest free nilpotent group. However, I do not see how to prove it for matrices over ${\rm Heis}_3(\mathbb Z)$, nor for general free nilpotent groups. Note that a proof for all free nilpotent groups would imply the result for residually free nilpotent groups, which include (I believe) most fundamental groups of closed $3$-manifolds, in particular all the hyperbolic ones.
