The example in your PS is probably related to the Fourier series of the "sawtooth function" $f : [-\pi, \pi) \to {\bf R}$, $f(t)=t/\pi$, when we regard $f$ as an element of $L^\infty({\bf T})\cong {\rm VN}({\bf Z})$. I don't recall the precise formula for the Fourier series, but it certainly gives something which does not belong to $\ell^1({\bf Z})$.

More generally: an integrable function $h$ on ${\bf T}$ whose Fourier series belongs to $\ell^1({\bf Z})$ must be equal a.e. to a continuous function, and so any element of $L^\infty({\bf T})\setminus C({\bf T})$ would also suffice. It is worth noting that there exist functions in the disc algebra (i.e. continuous on the closed unit disc and analytic in the interior) whose Taylor series do not converge absolutely on the unit circle, and these give elements of $C({\bf T})\setminus {\ell^1}({\bf Z})$ (where by abuse of notation I am identifying $\ell^1({\bf Z})$ with a space of functions on the circle).

Anyway, regarding your original question: if $G$ contains an element of infinite order then this gives an inclusion of groups ${\bf Z} \to G$ and hence (since we are dealing with discrete groups) an inclusion of von Neumann algebras ${\rm VN}({\bf Z}) \to {\rm VN}(G)$, so you can embed the counterexample for ${\rm VN}({\bf Z})$ to get a counterexample for ${\rm VN}(G)$.