 Solution

The electronic transition starts from $A_{1g}$, so the electric-dipole test reduces to whether the final electronic irreducible representation occurs among $A_{2u}+E_{1u}$. The excited $E_{1u}$ state matches the $(x,y)$ dipole components and is therefore electric-dipole allowed.

The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is electric-dipole forbidden at the equilibrium geometry. Both excited states are singlets, so the difference is spatial symmetry rather than the spin selection rule. Inversion parity alone does not decide the result: both final states are ungerade, but only the matching vector representation is allowed.