TITLE: Does the classifying space of monoids commute with wedge sum up to weak equivalence?
QUESTION [5 upvotes]: If $G$ and $H$ are groups, then the map $BG\vee BH\to B(G\ast H)$ is a weak equivalence by van Kampen's theorem. However the classifying space $BM$ of a monoid can have arbitrary homotopy type so higher homotopy groups are involved.
If $M$ and $N$ are monoids, is the map $BM\vee BN\to B(M\ast N)$ still a weak equivalence? If there are counterexamples, is there still some criterion that insures such a weak equivalence?

REPLY [3 votes]: 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?