This answer rather answers a question of Giles in the comments and is essentially an expansion of the comment to Giles' answer; the aim is to show that if a commutative ring $K$ (of any characteristic!) is so that $K[G]$ satisfies the zero divisor conjecture, then there is no element $x \in K[G]$ such that $x^n = 1$ for $n>1$, unless $x \in K e_G$ (an element supported solely at the identity element of $G$; i.e. the ring $K$ itself has an element of $n$-torsion).
So assume $x^n=1$ and $x \notin K e_G$, and take $n$ to be the smallest integer to do so. There are two cases:
Case 1: if $K$ is of characteristic $p \neq 0$ and $p$ divides $n$, $0=x^n-1=(x^{n/p}-1)^p$ (Frobenius map; i.e. the binomial coefficients of a prime $p$ are all divisible by $p$). It cannot happen that $x^{n/p}-1 =0$ (since it would contradict the minimality of $n$). This means that $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor. A contradiction.
Case 2: if $p$ does not divide $n$ or $K$ is of characteristic 0, consider the augmentation map $\epsilon:K[G] \to K$... if $x = \sum_{g \in G} x_g g$ then $\epsilon(x) = \sum_{g \in G} x_g$. Since $\epsilon$ is a ring homomorphism, $\epsilon(x)^n = 1$ (so $\epsilon(x)^{n-1}$ is a multiplicative inverse of $\epsilon(x)$). Consider the element $x' = x \epsilon(x)^{n-1}$ Note that $x' \neq 1$, since that would mean that $x \in K e_G$. Then $\epsilon(x') = \epsilon(x) \epsilon(x)^{n-1}=1$ and $x'^n= x^n \epsilon(x)^{n(n-1)}=1$ (this uses commutativity of the ring $K$). Then $0=x'^n-1=(x'-1)(x'^{n-1}+\cdots+x'+1)$. Since $\epsilon(x')=1$, one has that $\epsilon(x'^{n-1}+\cdots+x'+1) = \epsilon(x')^{n-1} + \cdots + \epsilon(x') + 1 =n \neq 0$ as either $n$ is not divisible by $p$ or $K$ has characteristic 0. In particular $x'^{n-1}+\cdots+x'+1 \neq 0$ and $x'-1 \neq 0$. But $(x'-1)(x'^{n-1}+\cdots+x'+1) =0$ form a zero divisor.
Remark 1: Note that if $K[G]$ has no zero divisors, then any element $x$ such that $x^n = x^k$ for $n \neq k$, is a torsion element. Indeed, $0 = x^n-x^k = x^k (x^{n-k} -1)$. Since $x^k \neq 0$, this means that $x^{n-k}=1$.
Remark 2: I guess this is all well-known, but for some reason I could only find references to the case $n=2$ (which is much easier since $x^2-1 = (x-1)(x+1)$ and $x \pm 1$ is never 0 [unless $x = \pm e_G$]).