This exercise is from Quantum Computation and Quantum Information. Suppose the equality holds, and that $|ABR\rangle=\sum_i\sqrt{\lambda_i}|i\rangle_{AB}|i\rangle_{R}$ (with $\lambda_i>0$) is a purification of $\rho^{AB}$. Then we have \begin{align*} \rho^{ABR}&=\sum_{ij}\sqrt{\lambda_i\lambda_j}|i\rangle\langle j|\otimes |i\rangle\langle j|,\\ \rho^{AR}&=\sum_{ij}\sqrt{\lambda_i\lambda_j}\operatorname{tr}_B(|i\rangle\langle j|)\otimes |i\rangle\langle j|,\\ \rho^R&=\sum_i\lambda_i|i\rangle \langle i|. \end{align*} The identity $S(A,B)=S(B)-S(A)$ is equivalent to $\rho^{AR}=\rho^A\otimes \rho^R$, thus \begin{align} \sum_{ij}\sqrt{\lambda_i\lambda_j}\operatorname{tr}_B(|i\rangle\langle j|)\otimes |i\rangle\langle j|=\sum_{ij}\lambda_i\lambda_j \rho_i^A\otimes|j\rangle\langle j|. \end{align} It follows that $\operatorname{tr}_B(|i\rangle \langle j|)=0$ for $i\neq j$.
For any fixed $j$, we have $\lambda_j(\sum_i\lambda_i\rho_i^A-\rho_j^A)=0$, so $\rho_j^A=\sum_i \lambda_i\rho_i^A$ is constant, which is stronger than the statement $\{\rho_j^A\}$ have a common basis in the exercise. Is this an error? The statement for $\{\rho_i^B\}$ also seems false as in the comments of this post Prove the equality conditions in the triangle inequality $S(A,B)\ge |S(A)-S(B)|$ for the von Neumann entropy . From the above there shouldn't be any restrictions on $\{\rho_i^B\}$.
