Let $S$ be a $\mathbb Z$-graded ring and $f \in S$ be a homogeneous element. Is the morphism $\mathrm{Spec}(S_{(f)}) \to \mathrm{Spec}(S_0)$ induced by the morphism of rings $S_0 \to S_{(f)}$ (which is $S \to S_f$ in degree $0$) an open embedding?
This is clear when $\deg f = 0$ since $S_{(f)} \simeq (S_0)_f$. I don't know what happens when $\deg f \neq 0$ in general. Here is a strange situation that can happen: let $S = \mathbb Z[t,t^{-1}]$ and set $f=2t$. The morphism $\mathbb Z[t,t^{-1}]_0 \to \mathbb Z[t,t^{-1}]_{(2t)}$ is the morphism $\mathbb Z \to \mathbb Z_2$, so the morphism $\mathrm{Spec}(\mathbb Z_2) \to \mathrm{Spec}(\mathbb Z)$ is still an open embedding, but it is not the identity even though $\deg f > 0$.
Short answer is no: let $S=k[x_{0},...,x_{n}]$ and $f=x_{0}$, then you get a map from a dense open set in $\mathbb{P}^{n}$ to a point $\text{Spec}(k)$, which cannot be injective as a topological map.