At least for
As Pavel says, any semisimple $g$ I think this is element lies in a pretty well-studied question: here's an outline of an argumentmaximal torus, where you can take any root. Lets assume that On the other hand, if $G$ g$ is simply connectedunipotent, as then it is in the by image of the theorem exponential map, so you can make sense of Steinberg $g^\lambda$ for any $\lambda \in \mathbb C$, so counterexamples must have nontrivial semisimple and unipotent parts.
Suppose $g=s.u=u.s$ is the centralizers Jordan decomposition of an element. Then $s^n$ and $u^n$ are semisimple elements and unipotent respectively, so they are connectedthe Jordan decomposition of $g^n$. Thus the existence of roots is compatible with Jordan decomposition.
Now take a semisimple element $s$ such that $Z_G(s)^0$ does not contain a central torus (so its center is finite) -- if $x \in G$ is semisimplesimply connected then in fact $Z_G(s)$ is connected, so I'll assume that. Now pick a regular unipotent element $u$ in $Z_G(s)$ and consider $x$ had g =s.u$. I want to claim $n$-th roots g$ is a counterexample. Indeed suppose for all each $n$, then the corresponding stabilizers of the n$ we have $h_n$ an $n$-th roots should stabilize (as they are connected root, and their dimension $h=s_nu_n$ is bounded) to some subgroup its Jordan decomposition. Then $Z$ containing s_n^n =s$ and $x$, u_n^n=u$, and an infinite set of "roots" of both $x$ would be central s_n$ and $u_n$ lie in $Z$, so Z_G(s)$. Then we see that $s_n$ centralizes $u$ for all $n$, but since $u$ is regular in $Z_G(s)$ and $s_n$ semisimple it follows that $s_n$ must lie in the centre of $Z$ should contain a nontrivial torus.
Hence if Z_G(s)$. But then taking, say, $n$ equal to the order of that centre (which is finite) we want get a counter-example we need contradiction, as $s_n^n$ must then be $1$.
Semisimple elements $x$ s$ such that $Z_G(x)$ has semisimple rank equal to that of $G$ (or maybe I mean it's identity component in general). Such elements Z_G(s)$ does not contain a central torus exist, but there are only finitely many conjugacy classes of them, as was essentially shown in the paper of Borel and de Siebenthal. In fact I think that paper establishes that there are $r+1$ such classes where $r$ is the rank of $G$, so this would give a negative answer for the exceptional groups also. I suspect that these might somehow be the only counterexamples?