Given a group $G$ we may consider its group ring $\mathbb C[G]$ consisting of all finitely supported functions $f\colon G\to\mathbb C$ with pointwise addition and convolution. Take $f,g\in\mathbb C[G]$ such that $f*g=1$. Does this imply that $g*f=1$?

If $G$ is abelian, its group ring is commutative, so the assertion holds. In the non-abelian case we have $f*g(x)=\sum_y f(xy^{-1})g(y)$, while $g*f(x)=\sum_y f(y^{-1}x)g(y)$, and this doesn't seem very helpful.

If $G$ is finite, $\dim_{\mathbb C} \mathbb C[G]= |G|<\infty$, and we may consider a linear operator $T\colon \mathbb C[G]\to\mathbb C[G]$ defined by $T(h) = f*h$. It is obviously surjective, and hence also injective. Now, the assertion follows from $T(g*f)=f=T(1)$.

What about infinite non-abelian groups? Is a general proof or a counterexample known?