For g in G, write g=g_{s}g_{u} as its Jordan decomposition into semisimple and unipotent parts. I claim that the closure of the conjugacy class of g contains an elliptic element if and only if g_{s} is elliptic.

Let us first suppose that g_{s} is not elliptic. Choose an embedding of G into GL_{n}(ℂ). Then by our assumption, g_{s} has an eigenvalue of norm greater than one, let λ be the absolute value of such an eigenvalue. Suppose for want of contradiction that the conjugacy class of g_{s} contained an elliptic element a in its closure. WLOG a is in the special unitary group SU_{n}. Let h be in the conjugacy class of g_{s}. Then h has an eigenvalue of absolute value λ. Letting v be an eigenvector, we see that |(h-a)v| is at least (λ-1)|v|, so |h-a|≥λ-1, a contradiction.

Now suppose that g_{s} is elliptic. We may replace G by the centraliser of g_{s} is G, which is also reductive. So WLOG, g_{s} is central in G. Now the Zariski closure of the group generated by g_{u} is a one-dimensional unipotent subgroup of G. Let E be a non-zero element in its lie algebra. This is a nilpotent element. Then by the Jacobson-Morozov theorem, we can extend E to a sl_{2} triple E,F,H in Lie(G). Now consider conjugation by elements of the form exp(tH) with t real. This shows that g_{s} is in the closure of the conjugacy class of g, and we're done.