Here's a thought. Let $\newcommand{\hT}{h\mathrm{Top}}\hT$ be the homotopy category of spaces. I'll write $K_p=K(Z/2,p)$ for an Eilenberg MacLane space; it represents the functor $H^p=H^p({-},Z/2)$, so $\hT(X,K_p)=H^pX$.

Let $\newcommand{\K}{\mathcal{K}}\K$ denote the full subcategory of $\hT$ consisting of finite products of the $K_p$s.

Let $\newcommand{\A}{\mathcal{A}}\A$ denote the category of *product preserving* functors $\newcommand{\Set}{\mathrm{Set}}\K\to \Set$. Let $\newcommand{\grSet}{\mathrm{grSet}}U:\A\to \grSet$ be the functor defined by evaluating a functor at the objects $K_p$ of $\K$, so $U(F)_p=F(K_p)$.

The category $\A$ is precisely the category of unstable algebras over the steenrod algebra!
You can think of an object $F$ in $\A$ as the collection of sets $U(F)_p=F(K_p)$, together with, for each "cohomology operation" $K_{p_1}\times\cdots \times K_{p_r}\to K_q$, a function $U(F)_{p_1}\times\cdots \times U(F)_{p_r}\to U(F)_q$, which satisfy all the identities they need to satisfy.

This shows the precise sense in which unstable algebras are an algebraic category.

The functor $\newcommand{\op}{\mathrm{op}}H^\*\colon \hT^\op\to \grSet$ factors tautologically through $U:\A\to \grSet$, via the functor $A\colon \hT^\op\to \A$ defined by
$$A(X)(\prod K_{p_i}) = \hT(X, \prod K_{p_i}).$$

We'd like to show the functor $A$ is universal in some sense. So lets suppose we have some other small category $\newcommand{\L}{\mathcal{L}}\L$, that we let $\newcommand{\B}{\mathcal{B}}\B$ denote the category of product preserving functors $\L\to \Set$, and that we have a collection of objects $L_p\in \L$, at which we can evaluate to get a forgetful functor $U'\colon \B\to \grSet$. And that furthermore there is a functor $B\colon \hT^\op\to \B$ and a natural isomorphism $U'\circ B\approx H^\*$. In other words, $B$ is another factorization of $H^\*$ through an algebraic category.

Can we construct a comparison functor $C\colon \A\to\B$, ideally so that $C\circ A\approx B$ and $U'\circ C\approx U$? I don't know ... I suspect you need some additional conditions in order to do this. However, there is an obvious candidate, given by the coend construction. Thus, if $F\in \A$, then perhaps we ought to have $C(F)\in \B$ be defined by the coend in $\B$
$$ F\otimes_{\K} (B\circ i^\op)=\mathrm{Cok}(\coprod_K F(K)\times B(i(K)) \leftleftarrows \coprod_{K\to K'} F(K)\times B(i(K')))$$
where $i\colon \K\to \hT$ is the inclusion functor. But now I'm stuck ...