Let $(X,\mathcal{O})$ be a locally ringed space. For $x \in X$ a closed point (i.e. $\{x\}$ is closed in $X$), let $\mathcal{O}_x$ denote the stalk of $\mathcal{O}$ at $x$; and define the "punctured stalk" $\mathcal{P}_x$ of $\mathcal{O}$ at $x$ as the directed colimit (in commutative rings) of $\mathcal{O}(V\setminus\{x\})$ where $V$ ranges over all open neighborhoods of $x$ in $X$.
My question is, if $\mathcal{O}_x$ is an integral domain, do we have $\mathcal{P}_x = \mathrm{Frac}(\mathcal{O}_x)$, the fraction field of $\mathcal{O}_x$?
More generally, if $\mathcal{O}_x$ is not an integral domain, then is $\mathcal{P}_x$ the localization of $\mathcal{O}_x$ obtained by inverting all nonzero elements of its unique maximal ideal (which also consists of all its non-units)?
Edit: added the requirement that $x$ must be a closed point. (Thank you to the comments for pointing out this necessity.)
No. Let $X = \{x\}$ be a one-point space, with sheaf $\mathcal{O}$ defined by $\mathcal{O}(X) = \mathbb{C}$.
Then $\mathcal{O}_x = \mathbb{C}$, but $\mathcal{P}_x = \mathcal{O}(\varnothing) = 0$ is not the fraction field of $\mathbb{C}$.