This is an extended comment, follow-up to Giles Gardam's answer and the comments therein. Recall that the group of nonzero scalars $K^\times$ is a central subgroup of $(KG)^\times$ (for each group $G$ and field $K$).
Proposition. Let $G$ be a group and $K$ a field, such that $\bar{K}G$ has no zero divisor. Then for every field $K$, the group $(KG)^\times/K^\times$ is torsion-free. In particular, 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. Let $A$ be a $K$-algebra (=unital associative $K$-algebra) with no zero divisors. If $P\in K[t]\smallsetminus\{0\}$ and $P(x)=0$ for $x\in A$ then $x$ is a scalar.
Proof: just consider the unital $K$-subalgebra generated by $x$. It is commutative, finite-dimensional over $K$, and has no nonzero divisor, hence is a field, and hence is 1-dimensional since $K$ is algebraically closed.
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 $P\in K[t]$, 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 $KG$.
Lemma 2 in turn, applied to $P(t)=t^n-\lambda$ for any fixed $\lambda\in K$, immediately implies the proposition.
Edit: the second sentence of the proposition can be restated as: if $K$ is a domain and $\overline{\mathrm{Frac}(K)}G$ has no zero divisors, then every torsion element in $(KG)^\times$ is a scalar. This is obtained in ARG's answer just assuming that $KG$ itself has no zero divisors.