Am I correct that localising $k[x,y]/(xy)$ at $S=\{x,x^2,\dots\}$ gives: $$(k[x,y]/(xy))_{x}\cong k[x,x^{-1},y]/(y)\cong k[x,x^{-1}]?$$ My only confusion is that $x\in k[x,y]/(xy)$ is a zero divisor, and yet its image in the localisation is not.
Probably this is reasonable, since the two axes scheme $\text{Spec}(K[x,y]/(xy))$ has affine open subscheme $$D(x)=\text{Spec}(k[x,x^{-1}])\cong \Bbb A^1_K\backslash\{0\},$$ which is just the $x$-axis remove the origin, which is integral, so has no zero divisors.
Similarly, let $X$ be a scheme and $U$ some open subset. Let $s\in\mathcal{O}_X(U)$ be a non-zero-divisor. Is it possible that $\rho^U_V(s)\in\mathcal{O}_X(V)$ is a zero divisor?