Lemma: Let $S=\bigoplus_{d\geq 0}S_d$ be a graded ring and $f,g\in S$ be elements of positive degree such that $D_{+}(g)\subset D_{+}(f)$. Then $f$ is invertible in $S_g$.
I attempted to say because $D(g)\cap Proj(S)=D_{+}(g)\subset D_{+}(f)=D(f)\cap Proj(S),$ then $D(g)\subset D(f)$ hence the result follows. But that is certainly not true. Then how could one argue instead that $f$ is invertible in $S_g$?
The statement follows from the following exercise:
Exercise: Let $I$ be a homogeneous ideal contained in $S_+$. Then, its radical is the homogeneous ideal: $$\sqrt{I} = S_+\cap\bigcap_{p \in V_{+}(I)}p $$ where $V_{+}(I)$ is the set of homogeneous prime ideals that contain $I$ and not the irrelevant $S_+$.
From this exercise, you can prove that if $V_+(I)\subset V_+(f)$ then $f\in \sqrt{I}$. In particular, setting $I = (g)$ in your case gives you that $f$ is invertible in $S_g$.
