If we assume the additivity axiom for all coproducts (not just countable ones), then a homology theory carries all filtered homotopy colimits to colimits of abelian groups.

Let $J$ be a filtered category and given a homology theory $h_*$, consider two families of functors $\mathsf{Top}^J \to \mathsf{Ab}$. One is $X \mapsto h_*(\mathrm{hocolim}_{j \in J} X_j)$ and the other is $X \mapsto \mathrm{colim}_{j \in J} h_* X_j$. Both these families of functors are "homology theories" in the sense that they satisfy all the same axioms as $h_*$ (using the fact that $J$ is filtered to show that the latter is exact).

The universal properties of (homotopy) colimits yield a transformation which is an isomorphism "on coefficients" which in this case means on every representable functor over $J$. It follows (by the usual argument you mention in your question) that this transformation is an isomorphism everywhere. In this case, you need to observe that every diagram of spaces over $J$ is equivalent to a projectively cofibrant diagram, i.e. one obtained by pushing out coproducts of $J(j, -) \times S^n \to J(j, -) \times D^{n+1}$ and then taking *countable* sequential colimits. Compactness of $S^n$ implies that countable colimits in the small object argument suffice. Now we proceed by induction using exactness to handle the pushouts, additivity to handle the coproducts and Hatcher's exercise to handle the colimit of a countable sequence.

**Added:** the argument of the previous paragraph in more detail. Given diagrams $A, B \colon J \to \mathsf{Top}$, a map $f \colon A \to B$, $j \in J$, $n \in \mathbb{N} \cup \{ -1 \}$, consider the set $C_{j,n} (f)$ of all commutative squares$\require{AMScd}$

\begin{CD}
J(j, -) \times S^n @>>> A \\
@VVV @VV{f}V \\
J(j, -) \times D^{n+1} @>>> B.
\end{CD}

Define $\bar B$ by pushing out $\bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times S^n \to \bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times D^{n+1}$ along $\bigsqcup_{j,n} C_{j,n} (f) \times J(j, -) \times S^n \to A$.

Now, given any diagram $X \in \mathsf{Top}^J$, define $X^{(-1)} = \varnothing$ and $f_{-1} \colon X^{(-1)} \to X$ as the unique map. Then apply the above construction to get $X^{(0)} = \bar X^{(-1)}$ and the induced map $f_0 \colon X^{(0)} \to X$. Proceed inductively to obtain a sequence of maps $X^{(-1)} \to X^{(0)} \to X^{(1)} \to \ldots$ and let $\tilde X$ denote its colimit which comes with an induced map $\tilde X \to X$. This latter map is a weak equivalence, even an object-wise acyclic Serre fibration. Indeed, in any a square

\begin{CD}
S^n @>>> \tilde X_j \\
@VVV @VVV \\
D^{n+1} @>>> X_j
\end{CD}

the top map factors through some $X_j^{(k)}$ since $S^n$ is compact and the resulting square has a lift since it corresponds by an adjunction to one of the squares above and a solution to that lifting problem was adjoined in the construction of $X^{(k+1)}$.

Now, assume that $H_*, K_* \colon \mathsf{Top}^J \to \mathsf{Ab}$ are homology theories with a transformation $\phi \colon H_* \to K_*$ that is an isomorphism on representables. We proceed to prove that it is an isomorphism everywhere in a few steps.

Step 1. $\phi$ is an isomorphism on $J(j, -) \times D^{n+1}$ by homotopy invariance.

Step 2. $\phi$ is an isomorphism on $J(j, -) \times S^n$ by the usual Mayer--Vietoris argument (decompose the sphere into two hemispheres).

Step 3. $\phi$ is an isomorphism on both $\bigsqcup_{j,n} C_{j,n} (f_k) \times J(j, -) \times S^n$ and $\bigsqcup_{j,n} C_{j,n} (f_k) \times J(j, -) \times D^{n+1}$ by additivity.

Step 4. $\phi$ is an isomorphism on all $X^{(k)}$ by induction on $k$ applying exactness each time.

Step 5. $\phi$ is an isomorphism on $\tilde X$ by the mapping telescope trick.

Step 6. $\phi$ is an isomorphism on $X$ by homotopy invariance.