Let $G$ be an infinite group and $x \in R[G]$ the group ring of $G$ over $R$. If $gx-x=0$ $\forall g \in G, g\neq 1$, then follows $x=0$.
I need a nice proof. I already have one dealing with the 'support of x'. But this just something I can imagine it to be true and nothing to write down.
Ok, supposing (as anon suggested above) that you are requiring that $\forall g \in G$ should holds $gx - x = 0$ then there's a direct proof.
We have that $x= \sum_{g' \in G} r_{g'} g'$, where the $r_{g'} \in R$ are all but a finite number equal to $0$.
By hypothesis for every $g \in G$ we have $$\sum_{g' \in G} r_{g'} gg' = gx = x = \sum_{g' \in G} r_{g'} g'$$ if $g g' = g''$ for some $g'' \in G$ this tells that $r_{g'} = r_{g''}$.
Because for every pair $g', g'' \in G$ there's a $g \in G$ such that $g g' = g''$ we can conclude that for every pair $g',g'' \in G$ we should have $r_{g'} = r_{g''}$ , but we have that all $r_g$ but a finite number a null (and the $r_g$ in the decomposition of $x$ are infinite since $G$ is infinite, so there must be a $r_g=0$), so they must all be null.