This is related to assertion made in Hatcher, Algebraic Topology, Chpt 3, Sec 1 and Chpt 2.
I will write down axioms for cohomology and axioms for homology is written in a similar fashion.
Cohomology theory for CW complexes is a sequence of contravariant functors with boundary map $\delta:H^n(A)\to H^{n+1}(X/A)$ where $(X,A)$ is a CW pair which implies $X/A$ makes sense as CW complex and it satisfies the following axioms.
(1) If $f,g$ are homotopic, then $f,g$ induces same map on cohomology.
(2) For each pair CW complex $(X,A)$, there is a induced long exact sequence by $...H^n(X/A)\to H^n(X)\to H^n(A)\to H^{n-1}(X/A)...$
(3) For wedge sums of $X_i$ CW complexes, $X=\vee X_i$ for canonical inclusion $X_i\to X$, $H^\star(X)\cong\prod_iH^\star(X_i)$.
$\textbf{Q:}$ For Eilenberg-Steenrod Axioms, I do recall that for disjoint spaces $X=\bigsqcup X_i$, I have $H^\star(X)\cong\prod_i H^\star(X_i)$. How did disjoint union follows from above axioms? Note that I do not even have $H^i(pt)=0$ for $i\neq 0$ here and neither can I say $H^i(\bigsqcup x_j)=\bigoplus_jH^i(x_j)$ for $x_j$ being points. So I do not even see how to apply $(2)$.
Hatcher's approach is somewhat unusual, though it has the advantage of being very transparent. Hatcher defines a reduced single space cohomology theory for unbased CW-complexes. In fact, the occuring CW-complexes do not have basepoints, and there are no relative groups $H^n(X,A)$ but only single space groups $H^n(Z)$. The theory is reduced because all $H^n(pt) = 0$ (consider the exact sequence of the pair $(pt,pt)$ as in Rylee Lyman's comment).
The Eilenberg-Steenrod axioms apply to theories defined on pairs. Therefore you have to associate to a cohomology theory in the sense of Hatcher a cohomology theory for CW-pairs. Note that this will be a generalized cohomology theory (which does not necessarily satisfy the dimension axiom). Given Hatcher's $H^n$, the usual approach is to define $\mathcal{H}^n(X,A) = H^n(X/A)$. For $A = \emptyset$ we interpret $X/A$ as $X^+$ = disjoint union of $X$ and an isolated point not in $X$. You can show that this in fact a cohomology theory. The excision axiom has to be interpreted carefully because we can only excise $U \subset A$ such that $cl(U) \subset int(A)$ and $(X \setminus U, A \setminus U)$ is again a CW-pair.
Now you have $\mathcal{H}^n(X) = \mathcal{H}^n(X,\emptyset) = H^n(X^+)$. For $X = \bigsqcup X_i$ you get $$\mathcal{H}^n(X) = H^n(X^+) = H^n(\bigvee X^+_i) \approx \prod H^n(X^+_i) = \prod \mathcal{H}^n(X_i) .$$