Here is another sort of constraint. I'll write $Sp(\mathcal C)$ instead of $Stab(\mathcal C)$.
Claim: If $\mathcal A \simeq Stab(\mathcal X)$ for a nontrivial [1] $\infty$-topos $\mathcal X$, then for any nontrivial localization $Spectra_L$ of $Spectra$, the localization $\mathcal A_L$ is a nontrivial localization of $\mathcal A$.
Let $\mathcal X$ be a nontrivial $\infty$-topos, and let $x^* : Spaces {}^\to_\leftarrow \mathcal X : x_*$ be the unique geometric morphism to $Spaces$. Let us contemplate the accessible left exact functor $\xi = x_* x^* : Spaces \to Spaces$ (the "shape" of $\mathcal X$). Note that
- $\xi$ preserves the initial object $\emptyset \in Spaces$;
This is because if the initial object $t^\ast \emptyset = \emptyset_{\mathcal X}$ had a global section $1 \to \emptyset$, then $\mathcal X$ would be trivial.
- If $X \to Y \leftarrow Z$ are maps in $Spaces$ with empty pullback, then $\xi X \to \xi Y \leftarrow \xi Z$ likewise has empty pullback.
This follows from (1) since $\xi$ is left exact. Therefore
- If $X \in Spaces$ is disconnected, then so is $\xi(X)$.
For we can take different connected components of $X$ in (2) above. It now follows that
- The functor $Sp(\xi) : Spectra \to Spectra$ induced by $\xi$ is conservative.
This functor is induced by applying $\xi$ levelwise to each $\Omega$-spectrum. To verify the claim, note that if $0 \neq E \in Spectra$, then $E_n$ is disonnected for some $n$ (where we think of $E$ as an $\Omega$-spectrum $(E_n)_{n \in \mathbb Z}$). By (3), $Sp(\xi)(E)_n = \xi(E_n)$ is disconnected, and hence $Sp(\xi)(E) \neq 0$. As an exact, zero-reflecting functor between stable categories, this implies that $Sp(\xi)$ is conservative.
It now follows that
Theorem: Let $\mathcal X$ be a nontrivial $\infty$-topos, and let $\xi^\ast : Spaces \to \mathcal X$ be the inclusion of constant objects. Then the induced functor $Sp(\xi^\ast) : Spectra \to Sp(\mathcal X)$ is conservative.
This follows from (4) since postcomposing $Sp(\xi_\ast)$ results in the conservative functor $Sp(\xi_\ast \xi^\ast)$.
Proof of Claim:
Note that there is an adjunction $Sp(\xi^\ast) \dashv Sp(\xi_\ast)$. If $Sp(\mathcal X) = Spectra_L \otimes Sp(\mathcal X)$, then the left adjoint $Sp(\xi^\ast) : Spectra \to Sp(\mathcal X)$ factors through $Spectra_L$, contradicting the Theorem.
[1] An $\infty$-topos $\mathcal X$ is "nontrivial" if it is not equivalent to the the terminal $\infty$-category, i.e. if the map from the initial object to the terminal object of $\mathcal X$ is not an equivalence.