 Solution

For an irreducible $V$, Schur lemma makes the central element $-I$ act by a scalar $\epsilon$. Since $(-I{)}^2=I$, $\epsilon =\pm 1$, and on $V\otimes V$ it acts by ${\epsilon }^2=1$. Thus $\{\pm I\}$ lies in the kernel and the action factors through $S{U}_2/\{\pm I\}$, as in central parity on SU2 tensor products.

This fails for reducible representations. Take $V=V_0\oplus V_1$. On the summand $V_0\otimes V_1\subset V\otimes V$, the element $-I$ acts as $-1$, so the tensor-square action does not factor through the quotient.