No, they don't have to be homotopically equivalent. In fact:

- There is a map of CW-complexes $X \to Y$ which is an isomorphism on (co)homology, full stop, for every generalized (co)homology theory $h$, multiplicative or not.
- This is, in fact, equivalent to the map $X \to Y$ being an isomorphism on integral homology.
- Systematic examples of such maps are given by plus constructions.
- However, if the spaces involved are connected and simply connected, such a map must be a homotopy equivalence by the homology Whitehead theorem.

This leads to a more subtle question, which is:

- Are there two simply-connected spaces whose cohomology rings are isomorphic for all generalized cohomology theories $h$, but which are not homotopy equivalent?

Then the answer is yes. The spaces $S^3 \vee S^5$ and $S^1 \wedge {\mathbb{CP}}^2$ have the same cohomology rings for all generalized cohomology theories, but they're not homotopy equivalent. This is the suspension of Neil Strickland's example in the other question, which has the effect of killing all the multiplication in the cohomology ring.

EDIT: In the new version, where cohomology operations are allowed, the question becomes more difficult because you're moving in the right direction: attaching more data like cohomology operations (the next to allow would be secondary operations). I think this example covers that case: you have two spaces, $S^3 \vee S^3 \vee S^8$ and another one formed by some Whitehead-product cell attachment, whose cohomology rings both have trivial multiplication. The cohomology operations on the Whitehead-product space are all trivial because the 8-cell is attached to both copies of $S^3$, but not to either one individually: specifically, if you collapse down either copy of $S^3$ to a point you're left with something homotopy equivalent to $S^3 \vee S^8$.