 Solution

The two independent columns of $V$ remain independent after multiplication by invertible $S$. Since $R=SV$ has zeros below the second row, its leading two-by-two triangular block is invertible. Hence the image under $S$ of the invariant subspace is exactly $\mathrm{span}(e_1,e_2)$. This coordinate plane is invariant under $\hat{A}=SA{S}^{-1}$, forcing the bottom-left block to vanish:The block determinant factors $\det (zI-\hat{A})=\det (zI-B)\det (zI-D)$. Therefore the eigenvalues of $A$ are exactly those of the indicated two-by-two and $(n-2)$-by-$(n-2)$ blocks, counted with algebraic multiplicity. No diagonalizability assumption is needed.