In this answer to What is the relationship between connective and nonconnective derived algebraic geometry? I learned that any quasicompact open subscheme of an affine scheme is affine in the sense of nonconnective spectral algebraic geometry. Let me try to give this argument in my own words: let $X = \operatorname{Spec} A$ for an ordinary commutative ring $A$ and let $i : U \subseteq X$ be quasi-compact. We can find a finite cover of $U$ by distinguished opens, i.e. $U = D(a_1) \cup D(a_2) \cup \dotsb \cup D(a_m)$. Then the koszul complex $K = K(a_1,\dotsc,a_m)$ is a perfect complex with support $X \setminus U$. The derived pushforward $\mathbb{R}f_* : D_{\text{qc}}(U) \to D_{\text{qc}}(X) = D(A)$ is fully faithful, with left adjoint $\mathbb{L} f^*$, and embeds $D_{\text{qc}}(U)$ as the right orthogonal complement of $K$. Since $\mathbb{L} f^*$ is strong monoidal its right adjoint acquires a lax monoidal structure, and so the localization functor $\mathbb{R}f_* \circ \mathbb{L}f^*$ is (symmetric) lax monoidal. The image of the base ring $A$ under this is then a (not-necessarily-connective) $E_{\infty}$-algebra $B$ over $A$, whose category of modules may be identified with $D(U)$. The underlying complex carrying the $E_{\infty}$-structure can be calculated as the cofiber of $\mathbb{R}\Gamma_I(A) \to A$. We could also calculate $B$ via the sheaf condition for the structure sheaf of $\operatorname{Spec} A$ on the cover $\{D(a_1),\dotsc,D(a_m)\}$.
Now it makes sense to me that the functor points and the tt-category of $B$ are correct, but I don't see how the spectrum actually gives the right topological space. I tried to work this out in the specific example of $\mathbb{A}^2_{\mathbb{C}} \setminus \{(0, 0)\} \subseteq \mathbb{A}^2_{\mathbb{C}}$ (see the other answer to the linked question) and got confused. In that case we have $\pi_0(B) \cong \mathbb{C}[x, y]$, and my understanding is that the spectrum of an $\mathbb{E}_{\infty}$-ring is just the spectrum of its $\pi_0$-ring as a topological space (and then we equip it with a certain structure sheaf using the higher information). So in this case it would seem that the underlying/truncated ordinary schemes of $\operatorname{Spec} B$ and $U$ have different points. What gives? Is there something sneaky going on which allows $\operatorname{Spec} B$ and the open subscheme $U$ to be isomorphic in the $(\infty, 1)$-category of nonconnective spectral schemes?
