Charles has given a very good answer to the question.
The following is not meant to be an answer, but just a heuristic argument which I cannot
make into a proof.
There should be an operation, "free product," denoted $\sharp$, in the category of associative topological monoids. If $X,Y$ are based spaces, then
$$\Omega (X \vee Y)$$ (Moore loops), should decompose (at least up to homotopy) in this category as
$$(\Omega X) \sharp (\Omega Y) .$$ The reason I find this to be plausible is that a loop in $X \vee Y$ is clearly a word of loops of $X$ and $Y$, and a word is supposed to represent an element of the free product.
Supposing this to be the case, we could take $X = BM$ and $Y = BN$, then we would have
\Omega (BM \vee BN) \simeq (\Omega BM) \sharp (\Omega BN)
It should also be the case that the inclusion
$$M \ast N \to (\Omega BM) \sharp (\Omega BN)$$
is group completion, since $ (\Omega BM)$ and $ (\Omega BN)$ are group-like and the operation $\sharp$ should preserve grouplike monoids (furtheremore, we also should have
$M \ast N \simeq M\sharp N$). If this is true, then $(\Omega BM) \sharp (\Omega BN) \simeq \Omega B (M\ast N)$.
If the above works, then the homomorphism
\Omega (BM \vee BN) \to (\Omega BM) \sharp (\Omega BN)
is an equivalence. Now apply the classifying space to get the desired equivalence
$BM \vee BN \simeq B(\Omega BM) \sharp (\Omega BN) \simeq B(M \ast N)$.
Question: Can this heuristic sketch be made into a proof?