Does Wolbert's derived equivalence between $E_*^R$-local $R$-modules and $R_E$-modules come from a Quillen equivalence? Let $R$ be a ring spectrum (in the world of EKMM $S$-modules) and let $E$ be a smashing $R$-module. Denote by $R_E$ the $E_*$-localization of $R$. By a theorem of Wolbert (Theorem 2 in Classifying modules over $K$-theory spectra), the derived category $\mathrm{Der}(R)[E^{-1}]$ of $E_*^R$-local $R$-modules is equivalent to the derived category $\mathrm{Der}(R_E)$ of $R_E$-modules. I wonder: is this equivalence induced by a Quillen equivalence
$$
L_E \mathrm{Mod}(R)\simeq_Q\mathrm{Mod}(R_E)\; ?
$$
 A: This is an elaboration on Lennart's comment.
This can be made to come from a Quillen equivalence.  Here are the ingredients you'd usually need to show it.  (Sorry, I don't have my copy of EKMM handy and so I can't provide theorem numbers.)  The problem is that you haven't specified $R_E$ as an actual object yet rather than just as a homotopy type.
There is a localization map $R \to R_E$.  This can be chosen as a map of $S$-algebras and with $R_E$ cofibrant as a right $R$-module (which is possible up to equivalence, but not automatically satisfied).  In this case, you can produce a Quillen equivalence.  If you've got a representative for $R_E$ that's not a cofibrant $R$-module, then you're typically going to have to replace it with an equivalent algebra $R'_E$ which is, and this will give a zigzag of Quillen equivalences $L_E Mod(R) \sim Mod(R'_E) \sim Mod(R_E)$.  (Really, all we'll need is a "flatness" property.)
Now let's give some details of the proof.
Unless I'm mistaken, the ordinary model structures on $Mod(R)$ and $Mod(R_E)$ are lifted from $Mod(S)$: a map is a weak equivalence or a fibration if and only if it is so in $S$-modules.  For this reason, the forgetful map is automatically a right Quillen functor.  Then $R_E \wedge_R (-)$ is a left Quillen functor, preserving cofibrations and weak equivalences.
Now let's talk about the localization.  This is usually taken to be a left Bousfield localization on the model category level: $L_E Mod(R)$ has the same underlying category, and the same underlying cofibrations, as $Mod(R)$, but new weak equivalences (the $E$-equivalences) and new fibrations.
Since it's the same underlying category, we still have the forgetful functor from $Mod(R_E)$ to $L_E Mod(R)$, with left adjoint $R_E \wedge_R (-)$.  The left adjoint still preserves cofibrations.
More, it preserves weak equivalences.  Since this is a smashing localization, a map $X \to Y$ is an $E$-equivalence if and only if it's an equivalence after taking derived smash product over $R$ with $R_E$.  Since $R_E$ is cofibrant, smash products represent derived smash products $\wedge^{\mathbb L}_R$.  Therefore, $X \to Y$ is an $E$-equivalence if and only if $R_E \wedge_R X \to R_E \wedge_R Y$ is a weak equivalence.
Now we finally need to show that this is a Quillen equivalence.  Suppose $X$ is a cofibrant $R$-module and $Y$ is a fibrant $R_E$-module.  We already showed that a map $X \to Y$ of $R$-modules is an $E$-equivalence if and only if the map $R_E \wedge_R X \to R_E \wedge_R Y$ is an equivalence.  The unit map $R_E \wedge_R Y \to Y$ is always an equivalence because $Y$ is $E$-local and this is a smashing localization.  Therefore, $X \to Y$ is an equivalence in $L_E Mod(R)$ if and only if its adjoint $R_E \wedge_R X \to R_E \wedge_R Y \to Y$ is an equivalence in $Mod(R_E)$, making this into a Quillen equivalence.
