Inverse of a set, possible? Just like ordinary algebraic operations have inverses, could we imagine the inverse of a set? Like $x\in\{x\}$ then maybe the inverse denoted $[|x|]$ would mean 
$$\{\ [|x|]\ \}=x$$
Would this idea ever be useful?
Also we could define the cardinality of the inverse of a set as for example $$\operatorname{card}[|x|]=-1$$
 A: No, that doesn't really work. The fundamental property we want the notation $\{x\}$ to have is that
$$ \forall y: y\in\{x\} \iff y=x $$
If we can't preserve that, then whatever we get is not worthy of being notated $\{x\}$.
Now consider your specification
$$ \{ [x] \} = x $$
and let's see where that leads us if we take for example $x=\{0,1\}$.
$$ \{ [\{0,1\}] \} = \{0,1\} $$
So by definition of $\{x\}$ we have
$$ \forall y: y \in \{0,1\} \iff y = [\{0,1\}] $$
In particular by setting $y$ to $0$ and then $1$ we get
$$ 0 = [\{0,1\}] = 1 $$
so $0=1$. And even worse, this reasoning doesn't depend at all on what $0$ and $1$ are, so the consequences of having your inverse is that it becomes impossible for a set to have two different elements. Either something must prevent us from forming $\{0,1\}$, or everything is equal to everything else. And neither of these is a useful situation for doing mathematics in.

Of course you can have your $[x]$ if you accept that it will only work as specified in the special case that $x$ is a set with exactly one element. In that case $[x]$ will of course be that element. But that's not very interesting either.
We can say, for example,
$$ (\exists y: x=\{y\}) \implies x = \{ \bigcup x \} $$
where $\bigcup x$ is the set of all elements of elements of $x$, whose existence is what the set-theoretic Axiom of Union guarantees. So in that restricted sense, $\bigcup x$ can be viewed as the "inverse" of $x$ you're envisaging.
A: This idea is used all the time in equivalence classes, e.g., $\mathbb{Z}_n$, where we send an integer $j$ to its equivalence class $[j]$, which is a set, a subset of $\mathbb{Z}$. This works for any set $S$ that is partitioned into subsets.
The key thing here is that $j \mapsto [j]$ is well defined.
