Trying to understand you correctly you're trying to be able to say in what sort of environment you are within the space, but not necessarily its exact location.
For example, consider the real numbers with the usual topology. Every two open intervals are isomorphic, so given a subset of $\mathbb{R}$ you want to say "This is an open interval." even if you cannot specify its exact endpoints.
What you have described in the italics is essentially taking all substructures of $L$ and classifying them by isomorphism classes. When teaching introduction to logic, we introduce the concept of isomorphism of structure in a first-order language, I will give the definition here, for simplicity I will minimize the language to have only one binary relation, one unary function and one constant symbol.
Suppose $L$ is a language with $R$, a binary relation; $F$, an unary function; and $c$ a constant.
Let $M,N$ be two structures for the language $L$. The function $f\colon M\to N$ is an isomorphism of structures if:
- $f$ is a bijection;
- $f(c^M)=c^N$ (that is $f$ translates the constants of the language);
- for every $a,b\in M$ the relation $\langle a,b\rangle\in R^M\iff \langle f(a),f(b)\rangle\in R^N$ (that is $f$ preserves the relation $R$); and
- for every $a\in M$ we have $f(F^M(a)) = F^n(f(a))$ (that is $f$ commutes with the function symbol $F$)
One could require only one way preservation of the interpretation, and add that $f^{-1}$ also have those properties.
In a way, this means that the way you interpret $L$ in the structure $M$ and in $N$ is exactly the same. Of course there can be many isomorphisms between two structures, even between a structure and itself there can be infinitely many.
This notion is a general way to view isomorphism of many structures in mathematics, be it linear spaces (by linear operators which are bijective), topological spaces (by homeomorphism), and so on.
Now, we also add the notion of substructure. Given the language has only relations in it, it makes things simple (as functions require slightly more work):
Suppose $L$ is a language without any function symbols and constants, and $M$ is a structure of $L$.
A substructure $N$ of $M$ (denoted usually $N < M$) is a structure of $L$ such that the universe of $N$ is a subset of the universe of $M$; and $R^N = R^M\cap |N|\times|N|$ (where $|N|$ is the universe of $N$).
If the language has constants and functions one requires that $N$ will also contain all the constants and will be closed under the functions, however in your case it is the simple case as $S$ is only made of relations.
Finally we reach the point, what it seems to me that you're doing in that italic text of yours is having a language which you call $S$ and a structure for it $L$, now you try and classify all the substructures of $L$ by isomorphism classes.
As far as the transformation group, I'm not sure I understand that part in your question, and I would love to add further information to my answer if you have any questions or clarifications for the question itself.
A final note that comes to mind is that sometimes when talking about subsets it requires second-order logic (namely being able to quantify subsets of the universe) but the ideas I've described above work about the same way.