Dold-Thom and infinite symmetric power of an $H$-space This question is pretty much this 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.
 A: 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 ${j+\ell}\choose {j}$.
