Consider that in a space S there are sets A and B, where B is the boundary of the compact and simply connected set A.
What assumptions are required to define a unique A (or "A like" set) with respect to B and S? The assumptions I have identified so far are:
B partitions S into exactly 2 non-empty sets,
S is unbounded, complete, and simply connected.
In other words, is there a more concise, or similar, way to say "Assuming B partitions S into exactly 2 sets, A is the one and only compact and simply connected set in the family of sets that are the closure of sets in the partition of S by B".
The scenario I have in mind is that S is the Euclidean 2-dimensional space $\mathbb{R}^2$ (or homeomorphic to $\mathbb{R}^2$), and B is a loop (i.e. a path that starts and ends at the same point) that doesn't intersect with itself.
OK. For "the example I have in mind," (a simple closed curve in the plane) the set you're looking for is called 'the interior of $B$', at least in some texts.
The existence and uniqueness of this interior was (early on in mathematics) assumed; then, with considerable effort, it was proved by Camille Jordan ca 1887. (The history of this proof is interesting in and of itself). By the way, it's one thing to prove that there's a set that $B$ bounds; it's another to prove that this set is homeomorphic to a disk. That's a separate theorem (Schoenflies theorem, 1906 -- almost 20 years later!).
To understand the subtlety of this question when you try to generalize, consider the analogous case of a sphere in 3-space. You'd like to say that a sphere in 3-space bounds a solid that's homeomorphic to a solid ball. That was proved, in a special case, by Alexander in (I think!) 1929 (another 20 years!); I'm pretty sure the paper was in the Proc. of the Nat'l Acad. of Science. A rather condensed version of Alexander's proof is on pages 1 and 2 of Hatcher's notes on 3-manifolds: https://pi.math.cornell.edu/~hatcher/3M/3M.pdf. On the other hand, if you want to allow full generality for the map from the 2-sphere to 3-space, then you cannot conclude that both "sides" of the sphere are "nice"; something called the Alexander Horned Sphere shows this.
Now suppose you vary the question a tiny bit, and allow the curve to cross itself (like a figure 8). [This is the "immersed case" rather than the "embedded case".]
You'd then think that it'd be obvious that the complement of the curve ($\Bbb R^2 - B$) would consist of several pieces, one of them unbounded ("the outside" or "exterior") and then some others, which seem as if they should all be topological disks. That's true, if you're willing to make some local smoothness assumptions. But what if you allow things like polygons? Then you could draw a 2-edge polygon (go from P to Q and then back to P) with no interior! If you insist on no "reflex vertices" (i.e., ones where the curve changes direction 180 degrees), and a finite number of vertices, then you might be able to show that an immersed no-reflex finite polygon bounds a finite collection of disk-like surface pieces. (I think Jordan's original proof, somewhat modified, might suffice.)
What about the analogous questions in 3-space? If you have a nice smooth "sphere-like" surface in 3-space, but it's possibly self-intersecting (i.e. immersed), does its complement have at least two components, one of them unbounded? What about the case of an immersed polyhedron (with no reflex edges, to avoid degenerate cases)?
Well, the complement does have an unbounded component in both cases, by a compactness argument. And in the smooth case, the complement has at least one bounded component; I believe that was proved in a paper by Mark Feighn sometime in the 1980s or 1990s. (!)
For the polyhedral case, there need not be a bounded component of the complement. The so-called "house with two rooms" provides an example.
Summary: in the rare cases when you can prove that such a set exists, it's been called the "interior of $B$", and it may (if you're lucky) actually be homeomorphic to a disk. For more complicated spaces (and I have no idea what your "open space $S$" might mean!) there's generally such a broad collection of challenges (the complement might have multiple disks, all of whose boundaries are $B$; the complement might contain no disks at all; the curve $B$ itself might not be contractible; ...) that trying to define "the interior" is like trying to define the notion of "the person I met today": about 99% of the time, it makes no sense at all, because you met multiple people or no one at all.
Anyhow...start with "Jordan Curve Theorem" (and Schoenflies theorem), read at least one full proof of each, and by then you'll probably understand why the general notion you're asking about hasn't got a name or been considered much (hence isn't very searchable).