The other major class of cases where one can construct $E$ is when $R$ is a localised regular quotient of $L$, or in other words $R=L[S^{-1}]/I$, where the ideal $I$ can be generated by a regular sequence. There is a long history of general results of this type. I think that the sharpest versions are in my papers "Products on $MU$-modules" and "Realising formal groups" (although the methods used are not so different from earlier work). Note that the latter paper works with the periodic spectrum $MP=\bigvee_{n\in\mathbb{Z}}\Sigma^{2n}MU$ rather than $MU$ itself; this is more natural for many purposes. Note that in this context we focus on $\pi_0$ and note that $\pi_0(MP)=L$.
There is one more useful construction which is less often discussed in the literature. Choose generators $a_1,a_2,\dotsc$ for $L=\pi_0(MP)$. Let $A$ be the monoid (under multiplication) generated by these elements, and put $T=\Sigma^\infty_+A$, or equivalently $T=\bigvee_{a\in A}S^0$. Now $\pi_*(T)=\pi_*(S)\otimes L$, and there is an evident map $f\colon T\to MP$ of naive ring spectra, which is an isomorphism on $\pi_0$. If we want to make an $MP$-algebra $E$, we could hope to start by making a $T$-algebra $E'$, and then put $E=E'\wedge_TMP$. In particular, suppose that $I$ is an ideal in $L$ that is generated by some subset of monomials in the generators $a_i$. Then it is easy to construct $T/I$ as $\bigvee_{b\in B}S^0$ for a suitable subset $B\subseteq A$, and we can hope to construct $MP/I$ as $T/I\wedge_TMP$. In particular, we can choose lifts in $\pi_0(MP)$ of the chromatic generators $v_k$, and use these as some of the integral generators $a_i$; then we can form $MP/(v_1^2,v_1v_2,v_2^2)$, which is similar but not identical to the thing that @LennartMeier mentioned in his comment.
However, to make this work, we need more highly structured versions of $T$, $MP$ and $f$ (because there is no general construction of smash products of modules over unstructured ring spectra). We can construct strictly commutative versions of $T$ and $MP$, and also of $T/I$ when $I$ is generated by monomials, but unfortunately $f\colon T\to MP$ cannot be a map of strictly commutative rings, because $\pi_0(MP)$ has interesting power operations and $\pi_0(T)$ does not. There are some pitfalls with model category structures that can cause trouble here, and I have not checked all the details, but I think one can do the following, for example in the category of EKMM spectra.
- In the EKMM context, the $0$-sphere spectrum $S$ is not cofibrant. We write $C$ for the cofibrant replacement, and $C^m$ for the $m$-fold smash power.
- Let $A_i$ denote the submonoid of $A$ generated by $a_i$ and put $T_i=\Sigma^\infty_+A_i=\bigvee_mS$. The spectrum $T=\Sigma^\infty_+A$ is then the smash product of all the $T_i$ (by which I mean, the colimit over $n$ of $\bigwedge_{i=1}^nT_i$).
- Let $T_i^c$ denote the cofibrant replacement of $T_i$, in the category of strictly commutative ring spectra. This has no simple description. Let $T^c$ denote the smash product of the $T_i^c$; I think this is the cofibrant replacement of $T$ in the commutative category.
- Put $T_i^a=\bigvee_mC^m$. This is the free associative ring generated by $C$, and is the cofibrant replacement of $T_i$ in the associative category. There is a natural weak equivalence $T_i^a\to T_i^c$ of associative rings.
- Let $T^a$ be the smash product of all the objects $T^a_i$. This is an associative ring, but is not cofibrant as such. I think that does not matter.
- Now take a strictly commutative model of $MP$. Using the freeness property of $T_i^a$ we get a map $f^a_i\colon T^a_i\to MP$ carrying the generator to $a_i$. As $MP$ is commutative, we can smash these together and pass to a colimit to get a map $T^a\to MP$ of strictly associative ring spectra. This gives us an associative $T^a$-algebra. To be on the safe side, we should probably take the cofibrant replacement of this in the category of $T^a$-algebras, to get an object $M^a$. We can then take $M^c=T^c\wedge_{T^a}M^a$. This is an associative $T^c$-algebra, and I think we have done enough cofibrant replacement to ensure that the underlying homotopy type is still $MP$.
- Now let $I$ be an ideal in $L$ that is generated by monomials, and let $B$ denote the set of monomials that do not lie in $I$. There is a fairly direct way to make $\Sigma^\infty_+B$ into a strictly commutative ring spectrum, and we let $T^c/I$ denote the cofibrant replacement. This has a natural map from $T^c$. Now we can define $MP/I=T^c/I\wedge_TM^c$.
The following is a communal answer from the algebraic topology Discord [1], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me).
Kiran suggested it be recorded here to ease future reference.
The idea is to produce two topological realizations $M$, $N$ of a single $MU_*$–module by finding two distinct resolutions whose effect on homotopy is the same.
The two associated square-zero extensions then give a counterexample.
We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku_*$–module structure to trivialize.
$\DeclareMathOperator{\Sq}{Sq}
\newcommand{\F}{\mathbb{F}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\HFtwo}{H\F_2}
\newcommand{\Susp}{\Sigma}
\newcommand{\co}{\colon\thinspace}$
The nice homotopy groups of complex $K$–theory, $ku_* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa_{ku}$ is $ku$–linear: in the diagram
$$
\begin{array}{ccccc}
& & \Susp^4 ku \\
& & u \downarrow \\
\Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\
& & u \downarrow \\
& & ku & \to & H\Z,
\end{array}
$$
the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear.
Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$).
Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$
The bottom $k$–invariant $\kappa_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q_2 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q_2 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$
Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F_2 \oplus \Susp^4 \F_2$, which splits as a $ku_*$–module — hence this $ku_*$–module could alternatively be modeled by $N = \Susp H\F_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map).
To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.
Best Answer
Put $R_*=\mathbb{Z}_{(p)}[x,y,y^{-1}]$ with $|x|=|y|=2$. Define $f,g\colon BP_*\to R_*$ by \begin{align*} f(v_i) &= \begin{cases} y^{p-1} & \text{ if } i = 1 \\ 0 & \text{ if } i > 1 \end{cases} \\ g(v_i) &= \begin{cases} x^{p-1} & \text{ if } i = 1 \\ y^{p^2-1} & \text{ if } i = 2 \\ 0 & \text{ if } i > 2 \end{cases} \\ \end{align*} This gives Landweber exact ring spectra $A_f$ and $A_g$ of heights one and two respectively, so for a finite spectrum $X$ of type $2$ we have $A_f\wedge X = 0 \neq A_g\wedge X$, so $A_f$ and $A_g$ are not equivalent as spectra.