$\newcommand{\Top}{\mathrm{Top}} \newcommand{\Grp}{\mathrm{Grp}} \newcommand{\Set}{\mathrm{Set}} \newcommand{\pinch}{\mathrm{pinch}}$
This is kind of an open-ended question but a possible answer was given by Wojowu in the comments: any functor $f:\Top_*\to \Grp$ which sends the point to $1$, the circle to $\mathbb Z$, is homotopy invariant and satisfies the Van Kampen theorem is isomorphic to $\pi_1$, at least when restricted to pointed connected finite CW-complexes (without the value at the point, the functor could be constant with value $\mathbb Z$).
The argument is as follows. First, $f$ is homotopy invariant and hence factors through the homotopy category of pointed topological spaces. By the Yoneda lemma applied in that category, it follows that picking out $1\in \mathbb Z\cong f(S^1)$ induces a map of set-valued functors $h:\pi_1\to f$. I claim that this is a group morphism. This is not completely trivial, and I will explain that later. In fact, it's not completely true: either $h$, or the morphism corresponding to $-1\in f(S^1)$ is a group morphism. This is what I will prove later.
For now, let's take it for granted. On $S^1$, $h$ sends $1\in\mathbb Z\cong \pi_1(S^1)$ to $1\in \mathbb Z\cong f(S^1)$ (or $-1$) and hence is an isomorphism. On disks, homotopy invariance shows that both $f$ and $\pi_1$ are trivial, and hence also on higher dimensional spheres by van Kampen. Since general finite pointed connected finite CW-complexes are built out of spheres by pushouts to which one can apply van Kampen, it follows that $h$ is an isomorphism on all of them, as was claimed.
To get to infinite CW complexes, one will need some kind of finitary-ness properties of $f$, such as the fact that it preserves filtered (homotopy) colimits/unions of CW-complexes. Then the same argument will work. To get beyond connectedness, one might simply impose it by asking that for any $(X,x)$, the inclusion of the component at $x$ induces an isomorphism on $f$.
It seems difficult to say anything for general topological spaces if you do not assume that the functor is invariant under weak homotopy equivalences.
Finally, the statement about $h:\pi_1\to f$ being a group map. Recall that the group structure on $\pi_1$ is induced by the cogroup (up to homotopy) structure on $S^1$. Unwinding in terms of the Yoneda lemma what it means for $h$ to be a group morphism, we find that it corresponds to the statement that the following two elements agree in $f(S^1\vee S^1)$: $f(i_a)(1) f(i_b)(1)$ and $f(\pinch)(1)$ where $i_a, i_b: S^1\to S^1\vee S^1$ are the two inclusions, and $\pinch: S^1\to S^1\vee S^1$ is the pinch map defining the multiplication on $\pi_1$.
By van Kampen, however, the map induced by $i_a$ and $i_b$, $f(S^1)*f(S^1)\to f(S^1\vee S^1)$ is an isomorphism, and so the comultiplication map on $S^1$ induces a cogroup structure on $f(S^1)$ in $\Grp$. The following is the key nontrivial statement:
Lemma: There are exactly two cogroup structures on $\mathbb Z$, corresponding to the morphisms $\mathbb Z\to \mathbb Z_a * \mathbb Z_b, 1\mapsto 1_a * 1_b$ and $1_b * 1_a$.
This lemma is a tedious if elementary verification with free words - it corresponds to the statement that the forgetful functor $\Grp\to \Set$ has exactly two group structures: the obvious one, or the opposite multiplication one. It was proved by Kan, and is also recorded in Comultiplications on Free Groups and Wedges of Circles (with other fun stuff about comultiplications in general)
From this, it is clear that either $f(i_a)(1) f(i_b)(1) = f(\pinch)(1)$, or $f(i_b)(1) f(i_a)(1) = f(\pinch)(1)$. In the latter case, picking $-1$ in place of $1$ in the definition of $h$ reduces us to the previous case, so that $h$ is indeed a group morphism, as was to be shown.
