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.


Let's assume as Tom Goodwillie mentions in the comments that the homotopy equivalence between $Sym^{\infty}(\mathbb{P}^{\infty})$ and $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ is induced by mapping $\mathbb{P}^{\infty}$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ where each map $\mathbb{P}^{\infty}\rightarrow K(2i, \mathbb{Z})$ is the map in homotopic to the one corresponding to the generator of $H^{2i}(\mathbb{P}^{\infty}, \mathbb{Z})$. Now the map from $\mathbb{P}^{\infty}$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ is extended to a map from $Sym^{\infty}(\mathbb{P}^{\infty})$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ by the $H$-space operation. So the final map is a map of $H$-spaces.



Dold-Thom and infinite symmetric power of an $H$-space
No, not quite.


Let's set the stage like this:


If $G$ is a connected $H$-space then $Sym^\infty G$ is a ring space, so that $Hom(X,Sym^\infty G)$ is also a ring space and $\pi\_\ast Hom(X,Sym^\infty G)$ is a graded ring.


Also, $Hom^0(X,G)$ is a connected $H$-space, so $Sym^\infty Hom^0(X,G)$ is a ring space and $\pi\_\ast Sym^\infty Hom^0(X,G)$ is a graded ring.


The canonical map $Sym^\infty Hom^0(X,G)\to Hom(X,Sym^\infty G)$ is a ring space map. Therefore $\pi\_\ast Sym^\infty Hom^0(X,G)\to \pi\_\ast Hom(X,Sym^\infty G)$ is a map of graded rings.


Now what does this have to do with cup products?


I would say that the graded ring $\pi\_\ast Hom(X,Sym^\infty G)$ is the cup product ring for cohomology of $X$ with coefficients in the graded ring $\pi\_\ast (Sym^\infty G)=H\_\ast (G)$.


That is, the multiplication
$$
H^i(X;H\_{2j}(\mathbb P^\infty))\times H^k(X;H\_{2\ell}(\mathbb P^\infty))\to H^{i+k}(X;H\_{2j+2\ell}(\mathbb P^\infty))
$$
is a cup product. But it's a cup product based on the multiplication
$$
(\*)\ \ \ H\_{2j} \mathbb P^\infty\times H\_{2\ell}\mathbb P^\infty\to H\_{2j+2\ell}\mathbb P^\infty
$$
(induced by the $H$-space structure). This cannot be identified with the cup product
$$
H^i(X;\mathbb Z)\times H^k(X;\mathbb Z)\to H^{i+k}(X;\mathbb Z),
$$
because the map $(\ast)$ above does not take generators to generators; there is a factor of ${j+\ell}\choose {j}$.


