Suppose $X$ is an integral scheme with generic point $\eta$, and $U \subset X$ is an open subscheme. If the restriction $\mathcal O_X(X) \to \mathcal O_X(U)$ is an isomorphism, do we necessarily have $U = X$? I believe this is true (I have an argument for the affine case) but I don't know how to approach the general case.
(The title of the question explains the motivation: the condition is equivalent to the injections $\mathcal O_X(X) \hookrightarrow \mathcal O_{X, \eta}$ and $\mathcal O_X(U) \hookrightarrow \mathcal O_{X, \eta}$ having the same image.)
Wrapping my previous comments (and then some) in an answer:
By Stacks 10.157.6, if $R$ is a normal Noetherian domain, $R=\bigcap_{\mathfrak{p}}{R_{\mathfrak{p}}}$ where $\mathfrak{p}$ runs through prime ideals of height one.
It follows that if $X$ is a normal integral Noetherian scheme, then for any closed subset $F\subset X$ with codimension at least $2$, $O(X) \rightarrow O(X \backslash F)$ is an isomorphism.
So open subsets are not determined by their ring of regular functions (as seen in the fraction field). But maybe points are determined by the image of their ring of stalks in the field of rational functions ?
Short answer: no.
Consider the line with two origins (glue two copies of $\mathbb{A}_1$ along $\mathbb{A}_1 \backslash \{0\}$): then the two origins (the zero in each copy) have, in the field of rational functions, the same ring of stalks. This is because $x=y$ in the field of rational functions ($x$ being the coordinate on the first copy of $\mathbb{A}^1$, $y$ the coordinate in the second copy).
Longer answer: actually, it’s yes for reasonable schemes.
Claim: let $X$ be a separated integral scheme, $x,y \in X$. If $O_{X,x}$ and $O_{Y,y}$ have the same image in $K(X)$, then $x=y$.
Proof: let $R$ be the image of $O_{X,x}$ in $K(X)$: it’s a local domain with fraction field $K(X)$. We have two maps $u,v: \mathrm{Spec}{R} \rightarrow X$ given by $\mathrm{Spec}{R} \cong \mathrm{Spec}O_{X,x} \rightarrow X$ (and same for $O_{Y,y}$).
There is a closed subscheme $Z$ of $\mathrm{Spec}{R}$ such that a morphism $t: S \rightarrow \mathrm{Spec}{R}$ factors through $Z$ iff $u \circ t=v \circ t$. Informally, $Z=\{z| u(z)=v(z)\}$; formally, $Z$ is the base change of the diagonal $X \rightarrow X \times_{\mathbb{Z}} X$ (a closed immersion as $X$ is separated) by $(u,v): \mathrm{Spec}{R} \rightarrow X\times_{\mathbb{Z}} X$.
The hypothesis means that the two morphisms $u \circ \eta$ and $v\circ \eta$ are equal, where $\eta$ is the map induced by the inclusion of $R$ in $K(X)$. So $Z$ contains the generic point of $\mathrm{Spec}{R}$. As $Z$ is a closed subscheme, $Z \rightarrow \mathrm{Spec}{R}$ is a surjective closed immersion, hence, as $R$ is reduced, an isomorphism, and $u=v$. Therefore, if $p \in \mathrm{Spec}{R}$ is the closed point, $x=u(p)=v(p)=y$.