 Solution

The Mumford rigidity lemma says that if $A$ is complete, $B$ is connected, and a morphism $F:A\times B\to Z$ maps the fiber over some $b_0\in B$ to a point, then $F$ is constant on every $A$-fiber and factors through $B$.

Put $y=f(e_X)$ and normalizeso $g(e_X)=e_G$. DefineWhen the first coordinate is $e_X$, this is constantly $e_G$. Apply rigidity with the second copy of the complete variety $X$ as the complete factor. It follows that $F(x,z)$ is independent of $z$, and evaluation at $z=e_X$ gives $F(x,z)=g(x)$. Thereforeso $g$ is a homomorphism of group varieties and $f=T_y\circ g$.