I think the answer is "if and only if the group $G$ is not cyclic". Why?

1) An element of $\mathbb C\left[G\right]$ is zero if and only if it acts as zero on each irreducible representation of $G$ (since $\mathbb C\left[G\right]$ is the direct sum of the endomorphism rings of the irreducible representations).

2) An element of $\mathbb C\left[G\right]$ acts on an irreducible representation of $G$ either as zero or as an automorphism (because each irreducible representation of $G$ is $1$-dimensional, since $G$ is abelian).

Hence, for a product of the form $\prod_{g\in S}\left(1-g\right)$ to be zero, where $S$ is some ordered list of elements of $G$, it is necessary and sufficient that for each irreducible representation of $G$, there exists some $g\in S$ such that $1-g$ acts as zero on the representation, i. e. that $g$ acts as identity on the representation. Applied to a list $S$ containing all elements of $G\setminus 1$ (maybe several times), this means that the product $\prod_{g\in S}\left(1-g\right)$ is zero if and only if no irreducible representation of $G$ is faithful. Easy manipulations with roots of unity show this to hold if and only if $G$ is not cyclic.