This question is pretty much [this](https://math.stackexchange.com/questions/4477619/a-map-from-the-symmetric-algebra-generated-by-the-first-cohomology-to-the-cohomo) question stated in slightly different way. All projective spaces are complex ones. Let's assume $X$ is a connected CW complex. We have a natural map in the following form:
$$f: Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty}))\rightarrow Sym^{\infty}(Hom(X, \mathbb{P}^{\infty}))\rightarrow Hom(X, Sym^{\infty}(\mathbb{P}^{\infty}))$$

Superscript zero denotes the connected component corresponding to zero. Since $\mathbb{P}^{\infty}\cong K(\mathbb{Z}, 2)$ so $\pi_0(Hom(X, \mathbb{P}^{\infty}))=H^2(X, \mathbb{Z})$. 

By Dold-Thom $\pi_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))=H_i(Hom^0(X, \mathbb{P}^{\infty})))$. There is a $H$-space structure on $Hom^0(X, \mathbb{P}^{\infty})$ induced from $\mathbb{P}^{\infty}$. This $H$-space structure turns $H_i(Hom^0(X, \mathbb{P}^{\infty})))$ into a graded algebra. Consequently $\pi_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))$ is a graded algebra. 

- Does $f$ map this algebra product to the cup product on the right side?

Note that homotopy groups of the right side is as the following:

We have $Sym^{\infty}(\mathbb{P}^{\infty})=\prod_{i=1}^{\infty}K(2i, \mathbb{Z})$ so $\pi_i(Hom(X, Sym^{\infty}(\mathbb{P}^{\infty})))=\bigoplus H^{\text{even}}(X, \mathbb{Z})$ if $i$ is even and $\bigoplus H^{\text{odd}}(X, \mathbb{Z})$ if $i$ is odd. So there is a natural cup product structure on the right.