Let $z \in C_G(S)$ and let $n \in N_G(S)$. We want to show that $nzn^{-1} \in C_G(S)$.
If $s \in S$, then $(nzn^{-1})s(nzn^{-1})^{-1}=nz(n^{-1}sn)z^{-1}n^{-1}$.
Now $n^{-1}sn \in S$ and so $nz(n^{-1}sn)z^{-1}n^{-1}=n(n^{-1}sn)n^{-1}=s$.
Firstly, any group with a nontrivial cyclic Sylow $2$-subgroup has a normal $2$-complement, so it cannot be simple. There is a very elementary proof of that. Look at the regular permutation representation. Then the image of a generator of a cyclic Sylow $2$-subgroup is an odd permutation, so intersecting with the alternating group gets you a subgroup of index $2$, and then you get the normal $2$-complement by induction.
Secondly, Burnside's Transfer Theorem states that if $P \in {\rm Syl}_p(G)$ and $P \le Z(N_G(P))$, then $G$ has a normal $p$-complement. So, in particular, (assuming $P$ nontrivial), $G$ cannot be simple. Note that this can only apply if $P$ is abelian.
In particular, if $G$ has an abelian Sylow $2$-subgroup with $N_G(P)=P$, then $G$ is not simple. This applies whenever ${\rm Aut}(P)$ is a $2$-group, which is the case when $P = C_4 \times C_2$. So if $G$ is simple and $|P| = 8$, then $P$ must be elementary abelian. In that case $|{\rm Aut}(P)|= 168$, so the possibilities for $|N_G(P)/C_G(P)|$ are $3$, $7$ and $21$.
In the ealier question Isaacs 5C10, you would still need to eliminate the case $|N_G(P)/C_G(P)|=3$. I am guessing that there are other results on transfer in Isaacs book that would enable you to do that (there is a result called the Focal Subgroup Theorem for example), but unfortunately I don't have Isaacs' book right now!
Best Answer
If $x\in C_G(H)$ then $xH=Hx$.
So $xHx^{-1}=H$.
Hence $x\in N_G(H)$.