= Solution
Use the <Holstein–Primakoff transformation> only on the physical occupation sector $0\le n\le2S$, with $n=a^\dagger a$ and $S^-=a^\dagger\sqrt{2S-n}$. Acting on number states gives
$$
S^+|n\rangle=\sqrt{n(2S-n+1)}|n-1\rangle,\quad S^-|n\rangle=\sqrt{(n+1)(2S-n)}|n+1\rangle.
$$
Consequently $S^+S^-=(n+1)(2S-n)$ and $S^-S^+=n(2S-n+1)$, so
$$
\boxed{[S^+,S^-]=2(S-n)=2S^z,\qquad[S^z,S^\pm]=\pm S^\pm.}
$$
The endpoint factors keep the spin operators inside the physical sector. The square-root ordering matters; replacing it by an unrestricted linear oscillator would not give the exact finite-spin algebra.
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