Kunneth Theorem and localisation I have two, somehow related questions, and I would be very grateful if you point out at some references if the answer is known. If these are too elementary or not research level, then please feel free to move it mathexchange. I am not an expert in this, so I thought it is worth asking here. 
According to Ravenel's orange book, the Nilpotence Theorem implies that $H\mathbb{F}$ and $K(n)$ are ``essentially'' the only homology theories for which the Kunneth isomorphism holds; I suppose this latter means that 
$$E_*(X\times Y)\simeq E_*(X)\otimes_{E_*(pt)}E_*(Y).$$
So, my first question is whether this conclusion is immediate or whether it is not that trivial and a proof is recorded somewhere? I have not tried to prove it.
Second, I wonder if there is a spectrum $E$ so that 
$$\pi_*L_E(S^0\wedge S^0)\simeq\pi_*L_ES^0\otimes\pi_*L_ES^0$$
where the tensor product is over some suitable coefficient ring, $L_E$ is the Bousfield localisation with respect to $E$, and $S^0$ is the sphere spectrum. Or, equivalently on the category of spaces 
$$\pi^s_*L_E(X\times Y)\simeq\pi^s_*L_EX\otimes\pi^s_*L_EY$$
again with tensor product over say $\pi_*L_ES^0$. A first guess is $E=H\mathbb{F}$ and $K(n)$, but these do not seem that immediate to me either. 

Edit. Perhaps it is better to replace $\times$ with $\wedge$, and ask for an isomorphism
$$\pi^s_*L_E(X\wedge Y)\simeq\pi^s_*L_EX\otimes_{\pi_*L_ES^0}\pi^s_*L_EY$$
as also in Neil's update to his answer.

The motivating example for the second question is the following. Consider a compact Lie group and an embedding $1^{\times n}<G^{\times n}$. Then, the transfer map of Becker-Schultz-Mann-Miller-Miller provides a transfer map (upon choosing a suitable twisting bundle) 
$$\Sigma^{n\dim(\mathfrak{g})}BG^{\wedge n}_+\to (B1^{\times n})_+=S^0.$$
Now, one may hope that in homotopy the image of 
$$\pi_*(\Sigma^{n\dim(\mathfrak{g})}BG^{\wedge n}_+)\to\pi_* B1^{\times n}_+=\pi_*S^0=\pi_*^s$$
falls into the submodule of decomposable elements in $\pi_*^s$. Here, $\mathfrak{g}$ is the Lie algebra of $G$. This, however, is not true, which I think is mainly because of the lack of a Kunneth theorem. So, one hope is to find some $E$ where homotopy of $E$-localised spaces/spectra admits some Kunneth theorem. For this reason, I am interested in a Kunneth theorem and not a Kunneth spectral sequence. 
Equivalently, if there is a way to describe the submodule of decomposable elements in $\pi_*^s$ then I would be happy.
 A: Jeff Strom correctly pointed to Hopkins-Smith II for the Kunneth Theorem question: the correct statement is that any E with a perfect Kunneth Theorem is (additively) a wedge of copies of $K(n)$ for some n, where $K(0) = H\mathbb Q$ and $K(\infty) = H\mathbb F_p$.  There is also a discussion of this in Hopkin's Durham 1985 proceeding paper, which also has, if I remember right, an influential appendix in which Mike is relating the closely related Thick Subcategory Theorem to classical ring theoretic settings. He was wondering if the `identifying the prime fields theorem" as above was equivalent to the Nilpotence Theorem, and not just a consequence.   
Regarding smashing the $n$--fold transfer: if $G$ contains the subgroup $\mathbb Z/p$, then this map has Adams filtration at least $n$, so the image in homotopy will be elements of Adams filtration at least $n$.  This is a decomposablity statement of sorts since a map has Adams filtration $n$ if it factors as $n$ maps each of which is zero in mod p homology. 
A: For the first statement, I think you want 
Prop. 1.8 in "Nilpotence and Periodicity II" by Hopkins and Smith.
A: 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.
UPDATE: for the second question, I do not believe that the natural map
$$ \pi_*(L_EX)\otimes_{\pi_*(L_ES)}\pi_*(L_EY) \to \pi_*L_E(X\wedge Y) $$
is a natural isomorphism unless $E$ is zero (in which case $L_E=0$) or $E=E\mathbb{Q}\neq 0$ (in which case $L_E$ is rationalization).  However, I cannot see a proof of this at the moment.
