Let $G$ be a commutative topological group. May in his textbook, A concise course in algebraic topology, gives a model of the classifying space $BG$ so that it is a commutative topological group itself. Knowing the standard group structure of $S^1$, I can image it is not hard to show that it is the same group structure obtained by regarding $S^1$ as $B \mathbb{Z}$ by checking some universal properties.
My question is how about $\mathbb{R}\mathbb{P}^\infty$, $\mathbb{C}\mathbb{P}^\infty$, or $BO$ and $BU$? The first two spaces are homotopic to $B (\mathbb{Z}/2)$ and $BS^1$, and we know that they admit group structures up to homotopy by the discussion above. The latter two are more complicated and one way to know that they admit group structures is by Bott periodicity. Is there a classical description for those? Can those group structures be made strict and how one can see it?
For $\mathbb{RP}^\infty$, there is the following explicit monoid structure. Consider it as the projectivization of the polynomial ring $\mathbb{R}[x]$. The multiplication of the polynomial ring then descends to a multiplication $\mathbb{RP}^\infty\times\mathbb{RP}^\infty\to\mathbb{RP}^\infty$. This makes $\mathbb{RP}^\infty$ a commutative topological monoid. It is not a group, but it has inverses up to homotopy since $\mathbb{RP}^\infty$ is connected. Similarly, $\mathbb{CP}^\infty$ can be made into a commutative topological monoid using the multiplication of $\mathbb{C}[x]$.
(In fact, if I remember correctly, the usual simplicial construction of the classifying spaces $B(\mathbb{Z}/2)$ and $BU(1)$ actually yields spaces that are literally homeomorphic to $\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$, which gives commutative topological group structures on them. I don't recall how to prove this though and I don't know of any nice geometric description of the resulting group operations. From the perspective of familiar topological groups like Lie groups, these group structures are pretty bizarre. In particular, for instance, every element of $B(\mathbb{Z}/2)$ satisfies $x^2=1$ since the same is true of $\mathbb{Z}/2$ and the group operation on $B(\mathbb{Z}/2)$ is obtained by applying the product-preserving $B$ functor to the group operation on $\mathbb{Z}/2$.)