= Solution
Write $M=\lambda(\Gamma)''$, the <group von Neumann algebra>, and define right translations by $\rho(g)\delta_h=\delta_{hg^{-1}}$. Left and right translations commute, so every $T\in M$ commutes with every $\rho(g)$. A <Von Neumann factor> means that the center of $M$ is exactly $\mathbb CI$.
Suppose every nonidentity <conjugacy class> is infinite and $T$ is central. Put $\eta=T\delta_e$. Since $T$ commutes with both regular actions and $\lambda(g)\rho(g)\delta_e=\delta_e$,
$$
\lambda(g)\rho(g)\eta=\eta.
$$
The action on a <basis> <vector> is $\delta_h\mapsto\delta_{ghg^{-1}}$, so the coefficients of $\eta\in\ell^2(\Gamma)$ are constant on conjugacy classes. A nonzero constant on an infinite class would have infinite squared sum. Thus all coefficients away from the identity vanish and $\eta=c\delta_e$.
Commutation with the right action now gives $T\delta_h=c\delta_h$ for every $h$, because those <vectors> form the right orbit of $\delta_e$. Hence $T=cI$ and $M$ is a factor. This argument also shows that $\delta_e$ is separating for $M$.
Conversely, if a nonidentity class $C$ is finite, then
$$
Z=\sum_{h\in C}\lambda(h)\in M
$$
commutes with every $\lambda(g)$, since conjugation permutes $C$. Therefore $Z\in M\cap\lambda(\Gamma)'=Z(M)$. It is not scalar: $Z\delta_e=\sum_{h\in C}\delta_h$ is nonzero and orthogonal to $\delta_e$. Consequently $M$ is not a factor. We have proved
$$
\boxed{\lambda(\Gamma)''\text{ is a factor}\iff\Gamma\text{ is an ICC group}.}
$$
The condition is vacuous for the trivial <group>, whose algebra $\mathbb CI$ is indeed a factor.
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