It is not really an answer to the question, but Ilya Bogdanov's answer puts quite strong restrictions on zero divisors of $\mathbb{F}G$ when $G$ is finite. Let $I(G)$ denote the augmentation ideal of $\mathbb{F}G$, that is $\{ \sum_{g \in G} \lambda_{g} g : \sum_{g \in G} \lambda_{g} = 0 \}$, which is indeed a two-sided ideal of $\mathbb{F}G$. Then every element of $I(G)$ is annihilated by $\sum_{ g \in G} g$, as Ilya observed, so $I(G)$ consists of zero divisors.

In the opposite direction, we can draw the conclusion that whenever $ab = 0$ for $a,b \in \mathbb{F}G$, at least one of $a$ or $b$ is in $I(G)$. For if $a$ is not in $
I(G)$, we can write
$ a = \lambda 1_{G} + i$ where $0 \neq \lambda \in \mathbb{F}$ and $i \in I(G)$.
Then $0 = ab = \lambda b +ib$. However, $ib \in I(G)$, so that $\lambda b \in I(G)$, and hence $b \in I(G)$. Similarly if $b$ is outside $I(G)$, we find that $a \in I(G)$, since $I(G)$ is a two-sided ideal. In particular, note that all nilpotent elements of $\mathbb{F}G$ lie in $I(G)$ by repeated applications of this.

(Later edit: In fact, note that a zero divisor $b \in \mathbb{F}G$ which annihilates
an element $a$ lying outside $I(G)$ must itself lie in $\cap_{n = 1}^{\infty}I(G)^{n}$, since we see above that there is some $i \in I(G)$ and non-zero $\lambda \in \mathbb{F}$ such that $ib = -\lambda b$ (and a similar argument if we had $ba = 0$).

3more comments