I think the converse holds. To prove the claim in the question, I attempted as follows:
$$ P(A, B) = P(A)P(B) $$ $$ \implies \sum_{C}{P(A, B, C)} = \sum_C{P(A, C)P(B)} $$ $$ \implies \sum_C{P(A, B, C)} = \sum_C{P(A)P(B)P(C)} $$
But that's as far as I could get. Does the claim in the question hold?
