(1) is trivial (the closure of an abelian group is abelian).
(2) follows because $C(S_w)$ is *the same* as $C(w)$, and it's much much easier to verify that $K_w = C(e^w)$ (where $w$ here is sufficiently small that the exponential map is a diffeo near it). Indeed, suppose $g$ commutes with $w$; then it commutes with $S_w$, and then with the closure. Now if $g$ stabilizes $e^w$, it commutes with $w$, and thus with $S_w$. Conversely if $g$ commutes with $S_w$, then clearly (by definition!) it commutes with $e^w$.