it would seem that it is sufficient for $ S$ to be be closed under applications of $ A^{-1}B$ and $ B^{-1}A$. Note $ A^{-1}B = (B^{-1}A)^{-1}$. $ A^{-1}B = \frac{(x+b)-x_0}{a}+x_0$ $ B^{-1}A = x_0+a(x-b-x_0)$ not sure if this is enough to characterize the sets, or if something more explicit is wanted, it certainly doesn't 'feel' complete.

That's pretty much a restatement of the problem. Is $ x_0$ fixed?

the problem reads as though $ A$ (and its variables $ a,x_0$) is fixed, and we need to find sets $ S$ for which there exists a $ B$ that satisfies the necessary closures. but it seems to me that its easier (and equivalent) to determine a collection of set $ S$'s for a fixed $ B$, correct me if my logic is flawed. what i should have said in my previous post was that for a fixed $ A$ and $ B$, then for any subset $ X \subset \mathbb{R}$ and take $ S$ to be the closure of $ X$ under the applications of $ A^{-1}B$ and $ B^{-1}A$. but the problem seems to suggest that there is a nice 'presentation' for such sets $ S$.

We claim that the superinvariant sets are $\varnothing$, $\mathbb{R}$, $\{p\}$, $\mathbb{R}\backslash\{p\}$, $(p,\infty)$, $[p,\infty)$, $(-\infty,p)$, $(-\infty,p]$. Clearly all these are superinvariant. Conversely, suppose $S$ is a superinvariant set. Case A: $S$ is periodic, i.e., there exists a period $t > 0$ such that $S+t = S$ (here $S+t$ means $\left\{s+t : s\in S\right\}$). If $t$ is a period, then scaling $S$ by a factor of $a$ shows that $at$ is also a period of $S$. Hence, either $S$ is empty, or else we can pick $s\in S$ and then deduce that $s-at, s + at\in S$ for all $a > 0$, and by varying $a$ we see that $S = \mathbb{R}$. Case B: $S$ is not periodic. We note that, in this case, for a given stretching $A$, we have a unique choice of the constant $b$ in the translation $B$. Now, let $a_1,a_2$ be positive reals not equal to $1$. We know there exist $b_1,b_2\in\mathbb{R}$ such that $S$ is invariant under the transformations $A_1(x) = a_1x-b_1$ and $A_2(x) = a_2x-b_2$. If we first apply $A_1$ and then $A_2$, we get the transformation $x\mapsto a_1a_2x - (a_2b_1 + b_2)$. If we first apply $A_2$ and then $A_1$, we get the transformation $x\mapsto a_1a_2x - (a_1b_2 + b_1)$. Both of these transformations are stretching by the same factor, and $S$ is invariant under both transformations, so by the afore-mentioned uniqueness we see that $a_1b_2 + b_1 = a_2b_1 + b_2$, i.e., $b_1/b_2 = (a_1 - 1)/(a_2 - 1)$. By varying $a_1$ and $a_2$ we can find a constant $p$ such that, for any $a > 0$, $S$ is invariant under $A(x) = ax - p(a-1)$, i.e., $A(x) = p+a(x-p)$. In other words, any stretching about $p$ preserves $S$. From this we deduce that: (i) if $S$ contains any point $q > p$, then $S$ contains the entire interval $(p,\infty)$. (ii) if $S$ contains any point $q <p$, then $S$ contains the entire interval $(-\infty,p)$. Combining these observations we may conclude that $S$ is a union of some of the following sets: $(-\infty,p)$, $\{p\}$, $(p,\infty)$, which gives us the answers.