Such multiplicative pairings on homology do often arise through cup product. However they usually not arise that naturally and also if so, it is harder to introduce them for example in terms of well-definedness. Maybe you want to look up the term "intersection form", which is dual to the cup product.
Note: Cohomology arises by applying $hom$ to the original chain complex, not on homology. It turns out, that there is always a relation between those two, but only in some cases cohomology arises by applying $hom$ to homology.
You are correct, that cohomology interpreted as a graded module/group might not give more information, regarding the resulting modules. However a functor also gives you maps, which is an additional tool which actually can give you additional information (if you have both), you can look up an example here.
Lastly, it might be interesting for you to see an example about why the graded ring structure might be important. For example, you would expect the cup product to behave natural under wedge sum. So consider $S^2 \vee S^4$ and also $\mathbb CP^2$. We have the same cohomology and homology graded module for both: $\mathbb Z,0,\mathbb Z , 0 ,\mathbb Z,0,0\cdots$. So there is no way to distinguish them so. But considering the graded multiplication structure on cohomology, it turns out that the generator of $H^2(S^2\vee S^4)= H^2(S^2) \oplus H^2(S^4)$ which arises from $S^2$ won't square to a non-trivial element in $H^4$, because this is generated by an element coming from $S^4$. So those (dimensions 2 and 4) are pretty independent, we basically superficially wedged together spheres to imitate $CP^2$'s cohomology module. However we have the case that a generator of $H^2(CP^2)$ squares to a generator of $H^4(CP^2)$. Now if we knew that cohomology is a functor to graded rings, i.e. multiplication behaves natural under maps, we know that those spaces can't be equivalent.