$\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained.

**Theorem 1:** Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups of $G$:

The image of $\bigoplus_F E_\ast(X_{hF}) \to E_\ast(X_{hG})$ contains all of the torsion;

The kernel of $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is contained in the subgroup of divisible elements.

This follows from the following two more precise theorems:

**Theorem 2:** Let $G$ be a compact Lie group, and let $X$ be a $G$-space. Let $N \subseteq G$ be the normalizer of a maximal torus $T \subseteq G$, and let $W = N / T$ be the Weyl group. Then $\Sigma^\infty_+ X_{hG}$ splits off of $\Sigma^\infty_+ X_{hN}$.

**Proof:** The splitting is given by the Becker-Gottlieb transfer: the fiber of $X_{hN} \to X_{hG}$ is $G/N$, the same as the fiber of $BN \to BG$, which has Euler characteristic 1.

**Theorem 3:**
Let $N$ be an extension of a finite group $W$ by a torus $T$, and let $E$ be a spectrum and $m \in \mathbb N_{\geq 2}$. Then the following hold, where $F$ ranges over finite subgroups of $N$:

$\varinjlim_F (E/m)_\ast(X_{hF}) \to (E/m)_\ast(X_{hN})$ is an isomorphism;

$(E/m)^\ast(X_{hN}) \to \varprojlim_F (E/m)^\ast(X_{hF})$ is an isomorphism.

**Proof of Theorem 1 from Theorems 2 and 3:** By Theorem 2, it will suffice to consider the case where $G = N$ is an extension of a finite group by a torus. Theorem 3 establishes Theorem 1 for $E/m$. Then (1) follows by considering the natural short exact sequence $0 \to E_{\ast}(-)/m \to (E/m)_\ast(-) \to E_{\ast-1}(-)^{\text{$m$-tor}} \to 0$, and the argument for (2) uses a similar exact sequence.

The proof of Theorem 3 will follow from a series of lemmas. For the remainder, we let $U(1)^n \to N \to W$ be an extension of a finite group by a torus, and we let $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ and $(C_q)^n \to N_q \to W$ be the subextensions which exist by the analysis in my other answer. We fix a spectrum $E$, $m \in \mathbb N_{\geq 2}$, and an $N$-space $X$.

**Lemma 4:** The fiber of $X_{hN_\ast} \to X_{hN}$ is $B\mathbb Q^n$, and in particular this map is an $(E/m)_\ast$ and $(E/m)^\ast$ equivalence.

**Proof:** This comes via a diagram chase from the fiber sequence $B\mathbb Q^n \to B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$.

**Lemma 5:** We have $(E/m)_\ast(X_{hN_\ast}) \cong \varinjlim_q (E/m)_\ast(X_{hN_q})$ canonically, and a canonical short exact sequence $0 \to \varprojlim^1 (E/m)^{\ast+1}(X_{hN_q}) \to (E/m)^\ast(X_{hN_\ast}) \to \varprojlim (E/m)^\ast(X_{hN_q}) \to 0$.

**Proof:** By the analysis in my other answer, we have $N_\ast = \varinjlim N_q$. Therefore $BN_\ast = \varinjlim BN_q$, and it follows that $X_{hN_\ast} = \varinjlim X_{hN_q}$. The lemma follows by the usual formulas for homology and cohomology of a filtered colimit.

**Lemma 6:** If $E$ is $m$-torsion, then $\varprojlim^1 E^\ast(X_{hN_q}) = 0$.

**Proof of Theorem 3:** This follows from Lemmas 4, 5, and 6, once we note that $E/m$ is $m^2$-torsion.

It remains to prove Lemma 6.

**Lemma 7:** Let $q,r \in \mathbb Z$, and consider the inclusion $C_{qr}^n \xrightarrow i (\mathbb Q/\mathbb Z)^n$. Consider also the inclusion $C_q^n \xrightarrow j C_{qr}^n$ with quotient $C_r^n$. Let $A$ be an $r$-torsion and $q$-torsion abelian group. Then $H^\ast(ij;A)$ is injective and $\Image(H^\ast(ij;A)) = \Image(H^\ast(j;A))$.

**Proof:** Direct calculation. More precisely, $H^\ast(BC_q;A)$ and $H^\ast(BC_{qr};A)$ both have $A$ in all degrees; the inclusion of $H^\ast(B(\mathbb Q/Z);A)$ is an isomorphism onto the even degrees, and $H^\ast(j;A)$ kills the odd classes while being an isomorphism on even classes. Then one extends this analysis to $n > 1$.

**Lemma 8:** Let $A \xrightarrow i B \xrightarrow j C$ be maps of chain complexes. Suppose that $ji$ is injective and $\Image(ji) = \Image(i)$. Then the sequence of homologies $H_\ast(A) \xrightarrow{i_\ast} H_\ast(B) \xrightarrow{j_\ast} H_\ast(C)$ has $i_\ast$ injective and $Image(j_\ast i_\ast) = Image(i_\ast)$.

**Proof:** Diagram chase.

**Corollary 9:** Fix $s \in \mathbb Z$, and consider the maps $H^s(BN_\ast;A) \xrightarrow i H^s(BN_{qr};A) \xrightarrow j H^s(BN_q;A)$. For $q$, $r$ sufficiently divisible by $m$ and $A$ $m$-torsion, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.

**Proof:** Using Lemma 7 as a base case, use Lemma 8 to induct through the pages of the Serre spectral sequences for the fibrations over $BW$. This is a first-quadrant spectral sequence, so for fixed $s$ it stabilizes at a finite page. The statement can be tested on associated gradeds, so there are no extension problems.

**Corollary 10:** Assume that $E$ is bounded below and $m$-torsion, and fix $s \in \mathbb Z$. Consider the maps $E^s(X_{hN_\ast}) \xrightarrow i E^s(X_{hN_{qr}}) \xrightarrow j E^s(X_{hN_q})$. For $q,r$ sufficiently divisible by $m$, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.

**Proof:** Using Corollary 9 as a base case, use Lemma 8 inductively to walk through the Atiyah–Hirzebruch spectral sequences for the fibrations over $BN_\ast$, $BN_{qr}$, and $BN_q$ respectively (which all have fiber $X$). Since we are assuming that $E$ is bounded below, this is essentially a first-quadrant spectral sequence so the argument goes in the same way as before.

**Proof of Lemma 6:** That this follows from Corollary 10 in the case where $E$ is bounded below can be seen in two ways — either from the eventual injectivity of $E^\ast(X_{hN_\ast}) \to E^\ast(X_{hN_q})$, or from the fact that sequence is Mittag–Leffler. When $E$ is not bounded below, we simply pass to a suitable connective cover of $E$, since we are always taking the cohomology of a suspension spectrum, which is bounded below.

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