Suppose that I have an isometry of $\mathbb{R}^{n}$, $\varphi(x) := A x + b$, where $A \in \mathrm{O}(n)$ and $b \in \mathbb{R}^{n}$.
If a set $S \subset \mathbb{R}^{n}$ satisfies $\varphi(S) \subset S$, does it then satisfy $\varphi(S) = S$?
This seems very reasonable to me, but I'm not sure how to prove it.
That's not true.
Look, e.g., at $n=1$ and $\phi(x) = x+1$. If $S=\{x: x\ge 0\}$, then $\phi(x) = \{x: x \ge 1\}\subset S$ (but not equal to $S$).
It is, very likely, true if $S$ is bounded, but that assumption you did not inlcude in your assumptions ;-)
Edit: I have to correct myself - there are counterexamples even in case that $S$ is bounded. It is easy to see though, that, in this case, $\varphi(S) \subset S$ will imply that $b=0$.
A counterexample must therefore be found by looking at the image of orthogonal maps and may be constructed as follows: let $0< \alpha < 1$ be any irrational number, and let $A$ be given as the rotation in two dimensional Euclidean space by $2\alpha \pi$. Then let $x_0=(0,1)$, and define recursively $x_n = A(x_{n-1})$. Then let $$S=\cup_{n\in \mathbb{N}}\{x_n\}$$ It's not difficult to see that then $\varphi(S)\subset S$, but $S \ni x_0\notin \varphi(S)$.
In the comments you will find a link for the case of compact $S$ (which would have been my next bet, otherwise).
No, and a counterexample with $n=1$ is provided by @Thomas above. However, exploring where the standard argument fails leads to a nice result.
First note that the inverse of an isometry always exists and is also an isometry: $$ y = Ax + b \quad\iff\quad x = A^{-1} (y - b) = A^{-1}y + (-A^{-1}b) $$ since $A^{-1} \in O(n)$ and $-A^{-1}b \in \mathbb{R}^n$.
Now, to establish the reverse inclusion $S \subseteq \varphi(S)$, take an arbitrary $y \in S$ and consider $x = \varphi^{-1}(y)$. Since $y = \varphi(x)$ by construction, clearly $y \in \varphi(S)$. The problem is that there's no reason to conclude in general that $x \in \varphi(S)$. We only know that $x \in \varphi^{-1}(S)$.
On way to recover a true result is to consider instead the group of isometries generated by by $\varphi$: $$ G = \langle \varphi \rangle = \{\dots, \varphi^{-2}, \varphi^{-1}, \operatorname{Id}, \varphi, \varphi^{2}, \dots\}. $$ Now it's true that if $G(S) \subseteq S$, then the reverse inclusion holds, hence $G(S) = S$, where $$ G(S) = \{g(s) \mid g \in G,\, s \in S\}. $$
