Na2MoO4 is mild oxidizer. At lower pH it will not react with Mn2+ (H2MoO4 + Mn2+ ---> 2 H+ + MoO2 + MnO2), also - will not oxidize water to O2 (H2MoO4 ----> MoO2 + 1/2 O2 + H2O), but this will not prevent formation of the precipitate.
What is needed is the reduction of Na2MoO4 to some soluble Mo3+ or Mo2+ salt (K4Mo2Cl8 is soluble but [Mo2Cl8]2- may still form precipitates with iron, copper etc.).