I am looking for references for the following facts. $\newcommand{A}{\mathbb{A}}\newcommand{P}{\mathbb{P}}\newcommand{Z}{\mathbb{Z}}\newcommand{Spec}{\mathrm{Spec}}\newcommand{sub}{\subset}\newcommand{O}{\mathscr{O}}$
Let $X$ be an integral scheme and $f \in R(X)$ a rational function with a representative $f : U \to \A^1_\Z$ for some open $U \sub X$. In the ring $R(X)$, $f$ is invertible if and only if the closure $\overline{f(U)}$ of its image does not contain the point of $\A^1_\Z = \Spec(\Z[T])$ corresponding to the prime ideal $(T) \sub \Z[T]$. In particular, if $f \in R(X)^*$, then the image $f(\eta)$ of the generic point of $X$ is not equal to $(T)$.
Suppose that $X$ is an $S$-scheme, where $S = \Spec(A)$. To $f$ corresponds a morphism $U \to \A^1_X$; let $g : U \to \P^1_S$ denote the composite with the projection $\A^1_X \to \A^1_S$ and the open immersion $\A^1_S \stackrel{\sim}{\to} D_+(T_0) \hookrightarrow \P^1_S = \mathrm{Proj}(A[T_0,T_1])$. Then the image $g(\eta)$ of the generic point is not contained in the union $V_+(T_0) \cup V_+(T_1)$ of the closed subschemes cut out by $T_0$ and $T_1$.
I have seen these results used implicitly in the Stacks project and other places, but I haven't found a proof (or even a proper statement) in either EGA or the Stacks project itself. I've written up proofs in detail here; they seem straightforward, but in my opinion not completely trivial (unless I'm overlooking something), and I'm surprised that the standard sources don't have them.