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$, so this would give a negative answer for the exceptional groups also. I suspect that these might somehow be the only counterexamples? 1 At least for semisimple$g$I think this is a pretty well-studied question: here's an outline of an argument. Lets assume that$G$is simply connected, as the by the theorem of Steinberg the centralizers of semisimple elements are connected. Now if$x \in G$is semisimple, and$x$had$n$-th roots for all$n$, then the corresponding stabilizers of the$n$-th roots should stabilize (as they are connected and their dimension is bounded) to some subgroup$Z$containing$x$, and an infinite set of "roots" of$x$would be central in$Z$, so the centre of$Z$should contain a nontrivial torus. Hence if we want a counter-example we need elements$x$such that$Z_G(x)$has semisimple rank equal to that of$G$(or maybe I mean it's identity component in general). Such elements exist, but there are only finitely many conjugacy classes of them, as was essentially shown in the paper of Borel and de Siebenthal. In fact I think that paper establishes that there are$r+1$such classes where$r$is the rank of$G\$, so this would give a negative answer for the exceptional groups also.