Imposing some reasonable conditions on our spaces (I think semilocally-simply-connected ought to do), one works through

Exercise 1 $\mathbb{Z}[X]$, the free topological $\mathbb{Z}$-module continuously generated by a convenient space $X$ is an $E^\infty$ space; the maps $\mathbb{Z}[X] \to \mathbb{Z}[Y]$ induced by $ X \to Y \to \mathbb{Z}[Y]$ make this construction continuously functorial; these induced maps are again $E^\infty$ maps.

Exercise 2 a weak homotopy equivalence of spaces $X \simeq X'$ induces a weak homotopy equivalence of $\mathbb{Z}$-modules.

Exercise 3 For a cofibration $X \to Y$, there is a pullback square $$ \begin{array}{c} \mathbb{Z}[X] & \to & \mathbb{Z}[Y] \\ \downarrow & & \downarrow \\ \mathbb{Z} & \to & \mathbb{Z}[Y/X] \end{array}$$

Exercise 4 $\pi_0 \mathbb{Z}[*] \simeq \mathbb{Z}$; otherwise $\pi_k \mathbb{Z}[*] \simeq 0$.

Exercise 5 the functor $X\mapsto \mathbb{Z}[X]$ preserves colimits of sequences of cofibrations.

Corollary We have verified that the functors $\pi_k \mathbb{Z}[X]$ satisfy the Eilenberg-Steenrod axioms for ordinary homology.

Again using the natural map $\mathbb{Z}[X] \to \mathbb{Z}$; write $\tilde{\mathbb{Z}}[X]$ for its kernel. To complete the exercises, მამუკა ჯიბლაძე's cogent remark explains why the natural map $SP^\infty X \to \tilde{\mathbb{Z}}[X]$ is an equivalence for connected $X$s.

But yes, it really is magical!

Imposing some reasonable conditions on our spaces (I think semilocally-simply-connected ought to do), one works through

Exercise 1 $\mathbb{Z}[X]$, the free topological $\mathbb{Z}$-module continuously generated by a convenient space $X$ is an $E^\infty$ space; the maps $\mathbb{Z}[X] \to \mathbb{Z}[Y]$ induced by $ X \to Y \to \mathbb{Z}[Y]$ make this construction continuously functorial; these induced maps are again $E^\infty$ maps.

Exercise 2 a weak homotopy equivalence of spaces $X \simeq X'$ induces a weak homotopy equivalence of $\mathbb{Z}$-modules.

Exercise 3 For a cofibration $X \to Y$, there is a pullback square $$ \begin{array}{c} \mathbb{Z}[X] & \to & \mathbb{Z}[Y] \\ \downarrow & & \downarrow \\ \mathbb{Z} & \to & \mathbb{Z}[Y/X] \end{array}$$

Exercise 4 $\pi_0 \mathbb{Z}[*] \simeq \mathbb{Z}$; otherwise $\pi_k \mathbb{Z}[*] \simeq 0$.

Exercise 5 the functor $X\mapsto \mathbb{Z}[X]$ preserves colimits of sequences of cofibrations.

Corollary We have verified that the functors $\pi_k \mathbb{Z}[X]$ satisfy the Eilenberg-Steenrod axioms for ordinary homology.

Using the natural map $\mathbb{Z}[X] \to \mathbb{Z}$, write $\tilde{\mathbb{Z}}[X]$ for its kernel. To complete the exercises, მამუკა ჯიბლაძე's cogent remark explains why the natural map $SP^\infty X \to \tilde{\mathbb{Z}}[X]$ is an equivalence for connected $X$.
Here is the remark: Some intuitive heuristics behind the fact are (a) $SP(X)$ is the free commutative topological monoid on $X$ (of sorts), (b) connected topological monoids possess homotopy inverses, so it is actually a free topological abelian group on $X$ (again sort of), (c) homology of $X$ is more or less the same as homotopy of the free topological abelian group on $X$. All this is almost rigorous in the simplicial context, where $\tilde H_*(X)=\pi_*\mathbb{Z}[X]$ more or less by definition.

But yes, it really is magical!