@TimCampion Well, there are a lot more affine non-connective spectral schemes than just of this form. Re: smoothness, the definition that works in the connective case is a locally fp morphism whose cotangent complex is of tor-amplitude $[0,0]$ (i.e. a retract of a free module of finite rank). With this definition, $V_X(F)$ would be smooth as a non-connective spectral scheme iff $F$ is of tor-amplitude $[0,0]$, so there are a lot of smooth connective spectral schemes which become non-smooth with this definition.

I think the issue is more fundamental. I'm saying that the embedding of SAG into nonconnective SAG is not compatible with many basic notions that I would call "geometric", like affineness, open immersions, smooth and étale morphisms... It's a genuinely different geometry, which makes the generalization from classical AG to connective SAG look very tame by comparison. So if you want to import some theorems from classical AG into SAG, you might have to be content with proving their connective versions!

Generally, stabilization is usually defined as a limit, using a right adjoint functor $v$ for the transition arrows in the tower. Then the answer to your question is yes, provided that the inclusion $F$ commutes with $v$, because limits of fully faithful functors are fully faithful. In the presentable case you can compute this limit as a colimit in the category of presentable $\infty$-categories and left adjoints (with the transition arrows in the tower being the left adjoint of $v$), but beware that this is not the same as the colimit in the category of $\infty$-categories.

Ah, good point. That simplifies the proof a bit as well, thanks!

Actually, the non-connective case then follows by taking connective covers (note that the map $R\{t\} \to R[t]$ is also compatible with formation of connective covers).

At least if $R$ is connective, the canonical map $R\{t\} \to R[t]$ is invertible iff $\pi_0(R)$ is a $\mathbf{Q}$-algebra. Since the map is compatible with extensions of scalars, it suffices to assume $R$ is discrete. Consider any residue field $R \to k$: if $k$ is of characteristic $p > 0$ then $k\{t\} \to k[t]$ cannot be invertible (since $\mathbf{F}_p\{t\} \to k\{t\}$ is faithfully flat, and you know $\mathbf{F}_p\{t\} \to \mathbf{F}_p[t]$ is not invertible), which means that $k$ must be of characteristic zero. Thus $R$ is purely of characteristic zero, i.e. a $\mathbf{Q}$-algebra.