Irrespective of the characteristic, it is not hard to prove that any $y \in FG$ for which $gy=y$ for all $g \in G$ is a scalar multiple of $x$. So the fixed submodule $\{ y \in FG : gy=y\,\forall g \in G\}$ of $FG$ is equal to $Fx$, and has dimension $1$.
$FG$ also has a submodule $I = \{ \sum_{g \in G} a_g g : a_g \in F, \sum_{g \in G} a_g = 0 \}$, which has codimension $1$ in $FG$, and $G$ acts trivially on $FG/I$.
If ${\rm char}\, F$ does not divide $|G|$, then $Fx$ and $I$ are disjoint and $FG = Fx \oplus I$.
But if ${\rm char}\, F$ divides $|G|$, then $Fx \le I$ and hence the module $FG$ has two distinct $1$-dimensional trivial composition factors, $Fx$ and $FG/I$. So if $FG$ were semisimple, then its fixed submodule would have dimension at least two, contradiction. So $FG$ is not semisimple as a module in this case.