Here is a family of examples. Let $F$ have characteristic two, and let $G$ be a finite group of order $2n$ with $n$ odd. Then $J(FG)$ squares to zero, and is therefore commutative.

Here is a family of examples. Let $F$ have characteristic two, and let $G$ be a finite group of order $2n$ with $n$ odd. Then $J(FG)$ squares to zero, and is therefore commutative.

Another (similar) family of examples is as follows. Let $F$ have characteristic $p$, let $H$ be a $p'$-group, and let $t$ be an automorphism of $H$ of order $p$ which fixes only the identity element. Let $G$ be the semidirect product $H \rtimes \langle t\rangle$. Then every non-principal block of $FG$ has defect zero, and the principal block is commutative, so $J(FG)$ is commutative.