Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$ for any $\operatorname{Spec} A$ with $x\in\operatorname{Spec} A$?

Yes that's true, the statement you want to prove is that if $U$ is any open subset of any ringed space $X$ then for $x\in U$ the natural map $\mathcal O_{U,x}\to\mathcal O_{X,x}$ is an isomorphism. It's pretty straightforward, for instance for surjectivity if you have an element $f_x\in\mathcal O_{X,x}$ then it is represented by a function $f\in\mathcal O_X(V)$ for some $V$ containing $x$, and then the stalk of $f|_{U\cap V}$ in $\mathcal O_{U,x}$ is the element you want mapping to $f_x$.