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Edit: Here's a related question. Does the comparison functor $\kappa : \mathsf{Mod}_A \to \operatorname{Desc}(\varphi)$ preserve compact objects? It does for discrete rings because it has a right adjoint given by a truncated cosimplicial limit, hence a right adjoint which preserves filtered colimits. But in the derived setting the limit is no longer truncated, so the same argument doesn't work. If this were true we could prove descent is equivalent to descent on compact objects (and get some nice criteria for effective descent too). This question came up when I was thinking of whether we could prove descent for pure morphisms using descent for split ring maps (fairly easy) plus the characterization of pure maps as those which are elementary equivalent to split ring maps. In fact we can, because we can write down first order axiom schema stating that the comparison functor is fully faithful on finitely presented modules (edit: this is not exactly right. If there is a counterexample to descent then we can capture it with finitely many parameters and a first order theory about them, but it's a bit subtle. I'm happy to go into detail if anyone is curious). But again this heavily uses that descent is about truncated/finite cosimplicial limits, because FOL is bad at talking about infinite things.

Edit 2: The answer to the question in the first edit is no. Take $\varphi : A \to k$ to be the projection of a Noetherian local ring onto its residue field. The comparison functor factors through the localization of $D(A)$ at the $k$-equivalences. By a result of Gunnar Carlsson, $k$-nilpotent completion agrees with the Greenlees-May derived completion; it follows that the map $\hat{A} \to k$ in the derived complete subcategory does satisfy descent. Since everything's presentable this means the associated comparison functor is a fully faithful left adjoint. The fully faithful left adjoint will reflect compactness (because coreflective subcategories are closed under colimits) and so if $\kappa(X)$ is compact in $\operatorname{Desc}(\varphi)$ then the derived completion of $X$ must be compact in the derived complete subcategory. Then by Greenlees-May duality $\Gamma_{\mathfrak{m}} X$ must be compact in $D^{\mathfrak{m}-\mathrm{tors}}(A)$. But again since $D^{\mathfrak{m}-\mathrm{tors}}(A)$ is coreflective its inclusion into $D(A)$ reflects compactness. In particular (assuming $\dim A > 0$) the unit $\Gamma_{\mathfrak{m}} A$ is not compact, so $\kappa(A)$ is not compact.

Edit 3: If $\varphi$ satisfies descent then it's a pushout-stable effective monomorphism in the category of commutative ring spectra (tensoring the amitsur complex of $B$ over $A$ with an $A$-algebra $C$ gives us the amitsur complex of $B \otimes_A C$ over $C$ because $- \otimes_A C : \operatorname{CAlg}(\mathsf{Mod}_A) \to \operatorname{CAlg}(\mathsf{Mod}_C)$ is a left adjoint). Is the converse true? It is for ordinary rings, because purity says the map is a pushout stable monomorphism (and effective monos are monic for $1$-categories). In analogy with the case of ordinary rings I wonder if we can characterize universal effective monos in terms of their Brown-Comenetz dual; it's known that pure maps in the sense of triangulated categories are exactly those whose Brown-Comenetz dual is a split epimorphism, and at this point my intuition is that that sense of purity isn't strong enough for descent (because it's testable on compact objects/"first order").

Edit 4: The answer to the question in edit 3 is yes. Every $A$-module $M$ is an $A$-linear retract of some commutative $A$-algebra, in particular the trivial square zero extension $A \oplus M$ (I am not so familiar with the topological cotangent complex setup but I believe this is still true for spectra). The class of $B$-nilpotent complete $A$-modules is closed under retracts because a retract of a limiting cone is still limiting. The assumption that $\varphi$ is universally an effective monomorphism says precisely that the underlying $A$-module of any $A$-algebra is $B$-nilpotent complete, and we've just seen that this implies every $A$-module is $B$-nilpotent complete, i.e. $\varphi$ is a descent map.

Edit 5: In the previous edit we said a map of ring spectra is a descent morphism iff it's a universal effective monomorphism. Dually a map of affine (nonconnective) spectral schemes is a descent map iff it's a universal effective epimorphism, which should(?) be the same as saying it's a cover in the canonical topology. So we're roughly asking if the codomain fibration defines a stack with respect to the canonical topology (when we know it defines a prestack). This seems very plausible but I don't really have a sense of whether it's true (in particular, infinite covers might be messy). On a related note it is true that any effective descent morphism for modules is also an effective descent morphism for the codomain fibration, because the functor $\mathrm{CAlg} : \mathsf{SymmMonCat} \to \mathsf{Cat}$ preserves limits (it is corepresentable). It's not clear to me that we can go the other way (effective descent for algebras => effective descent for modules) using the trivial square zero extension trick from before. It is of course true in the discrete case because we know descent for algebras <=> descent for modules and the latter is equivalent to effective descent.

Edit: The number of edits on this post was getting too long and so I've deleted them & summarize the main points here (see post history for more detail). The comparison functor $\kappa : \mathsf{Mod}_A \to \operatorname{Desc}(\varphi)$ does not preserve compact objects in the derived setting (unlike in the discrete setting); this and related facts like the limit expressing descent being infinite make me think purity in the sense of triangulated categories would not suffice to capture descent.

Note $\varphi$ is a descent morphism iff every $A$-module is $B$-nilpotent complete. Every $A$-module is $B$-nilpotent complete iff every $A$-algebra is $B$-nilpotent complete, because of the trivial square zero extension construction. So $\varphi$ is a descent morphism for modules iff it is a descent morphism for algebras (iff it is a universal effective monomorphism in the category of commutative ring spectra). If $\varphi$ is an effective descent morphism for modules then it is such for algebras because the functor $\operatorname{CAlg}(-)$ from symmetric monoidal $(\infty, 1)$-categories to $(\infty, 1)$-categories preserves limits (it is corepresentable in the $(\infty, 2)$-sense). The converse isn't clear to me but would follow if the formation of Beck modules preserved limits.

If that converse is true then we are then ultimately asking for a weak kind of Barr-exactness of $\mathsf{Aff} = \operatorname{CAlg}(\mathsf{Sp})^{\mathrm{op}}$. We want to know that the self-indexing/codomain fibration is a stack with respect to the topology generated by singleton covers consisting of an universal effective epimorphisms (we know it's a prestack, i.e. separated). This is strange because $\mathsf{Aff}$ is far from being regular, the only clear exactness property is has is extensivity, so maybe it's just a coincidence that things work out in the discrete case. None of the proofs I have seen of this fact in the discrete case seem to generalize well.

Edit 5: In the previous edit we said a map of ring spectra is a descent morphism iff it's a universal effective monomorphism. Dually a map of affine (nonconnective) spectral schemes is a descent map iff it's a universal effective epimorphism, which should(?) be the same as saying it's a cover in the canonical topology. So we're roughly asking if the codomain fibration defines a stack with respect to the canonical topology (when we know it defines a prestack). This seems very plausible but I don't really have a sense of whether it's true (in particular, infinite covers might be messy).

Edit 5: In the previous edit we said a map of ring spectra is a descent morphism iff it's a universal effective monomorphism. Dually a map of affine (nonconnective) spectral schemes is a descent map iff it's a universal effective epimorphism, which should(?) be the same as saying it's a cover in the canonical topology. So we're roughly asking if the codomain fibration defines a stack with respect to the canonical topology (when we know it defines a prestack). This seems very plausible but I don't really have a sense of whether it's true (in particular, infinite covers might be messy). On a related note it is true that any effective descent morphism for modules is also an effective descent morphism for the codomain fibration, because the functor $\mathrm{CAlg} : \mathsf{SymmMonCat} \to \mathsf{Cat}$ preserves limits (it is corepresentable). It's not clear to me that we can go the other way (effective descent for algebras => effective descent for modules) using the trivial square zero extension trick from before. It is of course true in the discrete case because we know descent for algebras <=> descent for modules and the latter is equivalent to effective descent.

Edit 5: In the previous edit we said a map of ring spectra is a descent morphism iff it's a universal effective monomorphism. Dually a map of affine (nonconnective) spectral schemes is a descent map iff it's a universal effective epimorphism, which should(?) be the same as saying it's a cover in the canonical topology. So we're roughly asking if the codomain fibration defines a stack with respect to the canonical topology (when we know it defines a prestack). This seems very plausible but I don't really have a sense of whether it's true (in particular, infinite covers might be messy).