= Solution
Expanding the total <angular momentum> and replacing transverse components by <ladder operators> gives
$$
\boxed{S_z=s_{1z}+s_{2z},\qquad S^2=s_1^2+s_2^2+2s_{1z}s_{2z}+s_{1+}s_{2-}+s_{1-}s_{2+}.}
$$
The <particle exchange operator> obeys $P_{12}^2=I$ and $P_{12}s_{1\alpha}P_{12}=s_{2\alpha}$, with the reverse relation for index $2$. These statements also apply to the <ladder operators>. Exchanging the indices preserves $s_1^2+s_2^2$ and $s_{1z}+s_{2z}$; it interchanges the last two terms in $S^2$. Operators on different tensor factors commute. Hence
$$
P_{12}S^2P_{12}=S^2,\qquad P_{12}S_zP_{12}=S_z,
\qquad\boxed{[P_{12},S^2]=[P_{12},S_z]=0.}
$$
The full particle exchange includes the <spin> coordinates as well as the spatial coordinates.
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