no, just take as a counterexmample: $V_1=U_1$, $V_2=U_2$, $\Sigma_A=\mathbb{1}$, $\Sigma_D=\mathbb{1}$, so $A+B=U_2(\Sigma_B+\mathbb{1})U_2^{\dagger}$, $C+D=U_1(\Sigma_C+\mathbb{1})U_1^{\dagger}$, so your assumption fails unless $U_1=U_2$.

no, just take as a counterexample: $V_1=U_1$, $V_2=U_2$, $\Sigma_A=\mathbb{1}$, $\Sigma_D=\mathbb{1}$, so $A+B=U_2(\Sigma_B+\mathbb{1})U_2^{\dagger}$, $C+D=U_1(\Sigma_C+\mathbb{1})U_1^{\dagger}$, so your assumption fails unless $U_1=U_2$.