Let $Y$ be integral (not necessarily separated) schemes, and let $y$ be a point of $Y$. Let $\xi_Y$ be the generic point of $Y$. The stalk $\mathcal{O_{\mathrm{Y},y}}$ can be canonically identified with a subring of the field $\mathcal{O_{\mathrm{Y}, \xi_Y}}$ via the ring homomorphism of direct limits induced from the homomorphism of directed systems from the directed system of sections of open sets containing $y$ to the directed system of sections over all open sets in $Y$ given by the identity on each ring.
If we have that under this identification, $\mathcal{O_{\mathrm{Y},y}} = \mathcal{O_{\mathrm{Y}, \xi_Y}}$ (note the equality under identification, and not just isomorphism), does it follow that $y = \xi_Y$?
One potential generalization of this question would be to ask whether, in an integral scheme, the stalk (under the canonical identification with a subring of $\mathcal{O_{\mathrm{Y}, \xi_Y}}$) determines the point uniquely. Since $Y$ is not assumed separated, I'm fairly confident that the example of two copies of $\operatorname{Spec} \mathbb{Z}$ glued together along the open subset $D(p)$ for some prime $p$ shows that in general, points of $Y$ need not be determined by their stalks – both of the "doubled" points will have stalk equal to the (unique) copy of $\mathbb{Z}[1/p]$ in $\mathbb{Q}$. I suspect that the same sort of counterexample cannot be created for the generic point, because you can't maintain integrality while "doubling" the generic point, but I don't see how to get from the algebraic statement ($\mathcal{O_{\mathrm{Y},y}} = \mathcal{O_{\mathrm{Y}, \xi_Y}}$) to the geometric statement $\overline{\{y\}} = Y$.
Best Answer
In our situation, $\mathcal O_{Y,y}$ is a field (and equal to $\kappa(y)$). So we get a morphism of integral schemes $\operatorname{Spec}(\mathcal O_{Y,y}) = \operatorname{Spec}(\kappa(y)) \to Y$ (Because we always have a canonical morphism $\operatorname{Spec}(\kappa(x)) \to X$ with image $x$ for a point $x$ in a scheme $X$)
This morphism is dominant, because it is injective on stalks. Hence it maps the generic point of $\operatorname{Spec}(\kappa(y))$ to the generic point of $Y$. But the image of the morphism is $y$. Thus $y=\xi_Y$. (See stacks for different conditions for dominant morphisms between integral schemes.)
(Note: We only used is that $\mathcal O_{Y,y}$ is a field, which is weaker a priori than being equal to $\mathcal O_{Y,\xi_Y}$)