The first statement is not true as stated. Firstly, if $E$ and $F$ have Künneth theorems then so does $E\times F$. (Here $E\times F$ is the same as $E\vee F$, but calling it $E\times F$ makes the ring structure more visible.) Secondly, suppose that $E$ has a Künneth theorem and $A_*$ is an algebra over $E_*$ that is flat as an $E_*$-module. Then the functor $X\mapsto A_*\otimes_{E_*}E_*(X)$ is a multiplicative homology theory that also has a Künneth theorem. By applying these constructions to the $K(n)$'s (including $K(0)=H\mathbb{Q}$ and $K(\infty)=H\mathbb{F}_p$) we obtain many examples. It would not surprise me if this gives all examples, but I do not think that this has been written down. One might need to worry about infinite products.
There is a related statement that is certainly true. Let $F$ be a ring spectrum such that $F_*$ is a graded field (ie every nonzero homogeneous element is invertible, and $F_*\neq 0$). This implies that every graded $F_*$-module is free, and thus that $F$ has a Künneth theorem. It also implies that $F\wedge X$ is always a wedge of (possibly suspended) copies of $F$. This includes the possibility that $F\wedge X$ could be zero, of course. This means that $F\wedge K(n)$ is a wedge of suspensions of $F$ and also a wedge of suspensions of $K(n)$. It follows from the Nilpotence Theorem that there exists a prime $p$ and a number $n\in [0,\infty]$ with $F\wedge K(n)\neq 0$. One can deduce from this that $F$ itself is a wedge of suspensions of $K(n)$. This still does not force $F$ to be isomorphic to $K(n)$, however; it could be $K(n)$ tensored with a finite field, or with a root of $v_n$ adjoined.