Let $R$ be a ring and let $I\subseteq R$ be an ideal of $R$. Then the Rees algebra of $R$ at $I$ is defined to be the graded ring
$$\operatorname{Bl}_I(R) = \bigoplus_{n\geq 0} I^n$$
Given an element $a\in I$, we can form the affine blowup algebra $R[I/a]$ to be the degree $0$ part of the graded localization of the Rees algebra at the element $a$ viewed as an element in degree $1$, $$R[I/a]:=(\operatorname{Bl}_I(R)_{a^{(1)}})_0$$
More explicitly, the elements of this ring are represented by elements of the form $\frac{x}{a^n}$ where $x\in I^n$, and they satisfy the usual equivalence relation $\frac{x}{a^n}\sim\frac{y}{a^m}$ if and only if there exists a natural number $k$ such that $a^k(a^m\cdot x - a^n\cdot y)=0$.
Suppose now that $I$ is a finitely generated ideal with generators $f_1,\dots,f_{n-1}$. We can always extend this family of generators to a family $f_1,\dots,f_{n-1},a$, since $a \in I \Rightarrow (a)+I=I$.
Let $\mathbb{Z}[X]:= \mathbb{Z}[x_1,\dots,x_n]$ be the polynomial ring in $n$ indeterminates over the integers. Specifying a family of $n$ elements of a ring yields a unique map $f:\mathbb{Z}[X]\to R$ such that $f(x_i)=f_i$.
Then form the affine blowup algebra $\mathbb{Z}[X][X/x_n]$. It is easy to see that this ring is isomorphic as a $\mathbb{Z}[X]$-algebra to $$\mathbb{Z}\left[\frac{x_1}{x_n},\dots,\frac{x_{n-1}}{x_n},x_n\right],$$ although abstractly as a commutative ring, this ring is again a polynomial ring in $n$ indeterminates.
(Note: We could equivalently describe the $\mathbb{Z}[X]$-algebra $\mathbb{Z}[X][X/x_n]$ by the endomorphism $\iota:\mathbb{Z}[X]\to \mathbb{Z}[X]$ sending $x_i\mapsto x_i \cdot x_n$ for $1\leq i <n$ and $x_n\mapsto x_n$.)
Then is it the case that $R[I/a]\cong \mathbb{Z}[X][\frac{X}{x_n}] \otimes_{\mathbb{Z}[X]} R$?
It seems like it ought to be, but I'm worried, since in the non-affine case, given a map $f:Y\to X$ and a closed subscheme $Z\hookrightarrow X$, we ordinarily need $Y\to X$ to be flat in order for the canonical map $\operatorname{Bl}_{f^{-1}(Z)}(Y)\to Y\times_X\operatorname{Bl}_Z(X)$ to be an isomorphism.
Tracked it down: It is true so long as $a$ is a non-zerodivisor in $R$.