$\newcommand\Spt{\mathit{Spt}}\newcommand\GrAb{\mathit{GrAb}}$Let $A$ be a ring spectrum. Suppose that $A$ has a Künneth theorem — i.e. the homology theory $A_\ast : \Spt \to \GrAb$ is a strong monoidal functor [1].
Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?
An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes Uniqueness of Morava $K$-theory.
[1] I have learned here that it's not so straightforward to say exactly what it means to "have a Künneth theorem" — right now I'm just assuming that there is some way to make $A_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.
Best Answer
If you want $K(h)_*$ to be strong monoidal, then you need the target category to be the category of graded $K(h)_*$-modules, not the category of graded abelian groups. Thus, I assume you are really asking for conditions under which $A_*(-)$ gives a strong symmetric monoidal functor to the category of graded $A_*$-modules.
Recall that the wedge of spectra is the same as the categorical product. It follows that the wedge also gives the categorical product in the category of homotopy commutative ring spectra. Also, it is easy to see that $\text{GrMod}_{A_*\times B_*}\simeq\text{GrMod}_{A_*}\times\text{GrMod}_{B_*}$ as symmetric monoidal categories. Thus, if $A$ and $B$ each have a Künneth isomorphism, so does $A\vee B$. In particular, any ring spectrum of the form $K(i)\vee K(j)$ has a Künneth isomorphism, but it is not a $K(h)$-module for any $h$ unless $i=j$.