This is an extended comment, follow-up to Giles Gardam's answer and the comments therein.
Proposition. Let $G$ be a group and $K$ a field, such that $\bar{K}G$ has no zero divisor. Then for every field $K$, every torsion element in $(KG)^\times$ is a scalar.
Here $K\subset \bar{K}$ is an algebraic closure. For arbitrary fields, the assumption on $G$ is satisfied by torsion-free virtually solvable groups (and hence residually torsion-free solvable groups, such as free groups and many more).
Lemma 1. Let $K$ be an algebraically closed field, if $A$ is a $K$-algebra (=unital associative $K$-algebra) with no zero divisor. If $P\in K[t]\smallsetminus\{0\}$ and $P(x)=0$ for $x\in A$ then $x$ is a scalar.
I argue by induction on the degree; degree zero is vacuously true. Suppose $P(x)=0$ with $P$ having positive degree $d$. Write formally $P(t)-P(u)=(t-u)W(t,u)$: then the degree of $W$ with respect to $t$ is $<d$ (and $W$ is nonzero). Since the degree is positive, $P$ has a root $s$ in $K$. So $0=P(x)-P(s)=(x-s)W(x,s)$. Write $W(t,s)=Q(t)$. If $x\neq s$, by the zero divisor fact, we deduce $Q(x)=0$. We conclude by induction that $x$ is a scalar, unless $Q=0$. (But $Q=0$ would mean $0=W(t,s)\in K[t]$, and hence $P(t)-P(s)=0$ as element of $K[t]$, and this means $P$ is constant, contradicting $d>0$.)
In turn this implies:
Lemma 2. Let $G$ be a group and $K$ a field for which $\bar{K}G$ has no zero divisor. Then for every nonzero polynomial $P$, every element $x$ in $KG$ such that $P(x)=0$ for some nonzero polynomial $P$, is a scalar.
Proof: by Lemma 1, $x$ is a scalar in $\bar{K}G$, and hence is a scalar in $KG$.
This in turn immediately implies the proposition.