If $(X,\mathcal O)$ is an affine variety over an alg.cl. field $k$, the stalk $\mathcal O_x$ at a point $x\in X$ can by constructed localizing the global section $\mathcal O (X)$ at the correct ideal.
(Example: If $x\in X=V(I)\in \mathbb A^n$, then $\mathcal O_x=\left(k[X]/I\right)_{\mathfrak m_x}$, where $V(\mathfrak m_x)=x$.)
More generally if $\mathcal O$ is a sheaf, the stalk at a point $x$ is defined as the direct limit $$\mathcal O_x:=\lim_{\substack{\rightarrow\\U\ni x}}\mathcal O(U),$$
and if $(X,\mathcal O)$ is a scheme, by definition, $\mathcal O_x$ is a local ring for all $x$.
If $X$ is projective, its global section $\mathcal O(X)$ is just $k$ and therefore $\mathcal O_x$ cannot be obtained localizing $\mathcal O_x$.
My question is: is there a somewhat canonical ring $\Omega$ that one can associate to a scheme $(X,\mathcal{O})$ so that each stalk $\mathcal O_x$ can be obtained localizing $\Omega$ at a certain ideal?
I think what you ask is not possible (or it depends what you mean by "canonical"). For example, if we assume that the set of maximal ideals of $\Omega$ is in bijection with points of $X$ (a natural assumption) we get
$$\Omega = \cap_{\mathfrak m} \Omega_{{\mathfrak m}} \cong \cap_{x \in X} \mathcal O_{X,x} = \mathcal O_X(X) $$
assuming that $X$ and $\Omega$ are noetherian. This is a contradiction if for example $X$ is projective.
On the other hand, for each affine cover $\{U_i\}_{i \in I}$ of $X$, let $\Omega = \prod_{i \in I }A_i$, with $\text{Spec}( A_i) = U_i$ so $\text{Spec}(\Omega) = \bigsqcup_{i \in I} U_i$. Now, let $x \in U_i$ and $\mathfrak m \subset \Omega$ the corresponding maximal ideal. Then by construction we have $\Omega_{\mathfrak m} \cong \mathcal O_{X,x}$. This procedure is not canonical at all because we needed to choose an affine open cover and for each $x \in X$ we needed to pick some $U_i$ with $x \in U_i$.