Let $G$ be a connected affine algebraic group over $\mathbb{C}$. It's a known fact that elements of $G$ admit a decomposition into semisimple and unipotent elements. Namely, choose a faithful representation $G \subseteq GL(n,\mathbb{C})$. Then given $g \in G$, the Jordan decomposition states that $g = su$, where $s$ is semisimple, $u$ is unipotent, and $su = us$. The crucial fact is then that $G$ contains both $s$ and $u$, and this decomposition is independent of the representation. Therefore, the decomposition $g = su$, as well as the notions of being semisimple and unipotent, are intrinsic to $G$. Furthermore, the decomposition is preserved by morphisms of algebraic groups.

However, if we allow holomorphic maps of Lie groups, the decomposition is not respected. For example, the exponential map $exp : \mathbb{C} \to \mathbb{C}^{*}$ sends unipotent elements to semisimple elements!

So this makes me wonder whether the decomposition depends on the algebraic structure of $G$.

For definiteness, here's my question: Let $G_{1}$ and $G_{2}$ be connected complex reductive groups over $\mathbb{C}$, and let $\phi: G_{1} \to G_{2}$ be a holomorphic group isomorphism. Does $\phi$ preserve the Jordan decomposition?

My feeling is that the answer is yes, and that it will come down to something like $G_{i}$ admitting a unique algebraic structure compatible with the group structure, and any holomorphic isomorphism being automatically algebraic.