I believe I finally found a proof. Surprisingly it does only use standard facts on Coxeter-groups (exchange condition, solving the word problem via braid-moves,...).

Let me first make the notation a bit easier:

Claim: Let $P\leq W$ be a special subgroup of $W$ generated by some subset $S'\subsetneq S$ of $S$ and $s \in S-S'$. Then the centralizer of $s$ in $P$ is generated by those involutions in $S'$ which commute with $s$, $C_P(s)=\langle s' \in S'~|~ s's=ss' \rangle$.

Proof: Let $w \in C_P(s)$ and $w=s_1...s_r$ be a reduced expression. By induction it is enough to prove that $s_rs=ss_r$ since the elements of length $1$ in the centralizer are precisely the simple involutions commuting with $s$.

 We have $\ell(ws)=\ell(w)+1$ and since $\ell(wsw^{-1})=\ell(s)=1$ we conclude that $\ell(wss_r)<\ell(ws)$, so $\ell(wss_r)=\ell(w)$. By the exchange condition there is a reduced expression for $ws$ ending in $s_r$ and since $s_1...s_rs$ is already a reduced expression for $ws$ there exists a finite series of braid-moves connecting these two expressions. 

 The expression $s_1...s_rs$ contains $s$ only once and no simple involution that does not commute with $s$ shows up to the right of $s$. Consider now any braid-move in this situation. If $s$ is not involved in the move the two conditions obviously still hold afterwards. If $s$ is involved the other simple involution involved must commute with $s$ since any braid-move involving $s$ and a non-commuting $s'$ requires either at least two occurrences of $s$ (to the left and to the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which happens. Hence any braid-move fixes our two conditions and after finitely many braid moves there is still no simple involution to the right of $s$ which does not commute with $s$. 

On the other hand there is, as noted above, a finite series of braid-moves after which the expression ends in $s_r$ so $s_r$ has to commute with $s$ as asserted.