Can one-element set be considered equal to its element? Are there "interesting" (that is non-trivial, for example not containing only one set) set theories with one element set being equal to their element ($\{x\}=x$ for every $x$)?
This question arose from the practical problem: Is it possible without "troubles" (such as contradictions) to consider an RDF term denoting a transformation between XML namespaces as an one-element set containing this term? If we can consider them equal, it makes shorter the notation, as we do not need to define an one-element set in this case but use the transformation term itself to denote this set. To define one term less in this case.
 A: Yes. If you take out one axiom (the Axiom of Regularity, which roughly disallows sets to be nested inside themselves), then ZFC is perfectly happy with the existence of sets $x$ such that $x = \{x\}$. Such sets $x$ are usually known as Quine atoms.
In fact there are many well-known set theories that explicitly allow the existence of Quine atoms, sometimes as a matter of principle --- New Foundations, for instance.
Of course it will never be the case for all $x$ that $x = \{x\}$. This would give a contradiction. Specifically, we can prove that $\varnothing \ne \{\varnothing\}$. If you wanted to change this, you would have to change the very definition of membership or equality of sets.
Given your intended computer programming application, I will also mention that there are programming languages which treat $x$ and $\{x\}$ as the same thing. This is not a problem, there is no contradiction, because "sets" in these languages are much more restricted than the sets of set theory. Typically, $\{\varnothing\}$ will not be an allowed set, and in general there is only a single level of nesting (i.e., no sets-of-sets).
A: Yes! ZFC - Regularity is consistent with as many Quine atoms you may add. However, to answer your question, matters largely depends on what you mean by saying "$\{x\}=x$ for every $x$"
If you mean:
$\forall x \exists y (y=\{x\} \land y=x)$
Then a model of this theory would be a nonempty set of Quine atoms, now we can have this set being infinite, such a set can be proven to exist in a version of ZFC-Regularity. A theory satisfiable in such a model would be one with axioms of Extensionality, Every set is a Quine atom, and for every $x_1,..,x_n$ objects there exists an object that is distinct from all of them. 
Such a theory is weak, but it satisfy this strong interpretation of your statement, and it is interesting according to the qualification you gave for being so.
If what you mean by "$\{x\}=x$ for every $x$", is:
$\forall x \forall y (y=\{x\} \to y=x)$
Then here matters totally differ since this doesn't entail that for every set there must exist a set that contains it as its sole element, so you can work with a theory in which only Quine atoms are its singleton sets! This can be done! Similar to the above, work in ZFC - Regularity + there exists infinitely many Quine atoms + there exists a set of all Quine atoms. 
Now  fix a set $U$ having $\aleph_0$ many Quine atoms. Define a special kind of power set operator $P^*$ as:$$P^*(x)= P(x) \setminus \{\{y\}| y \in x \land y \text { is not a singleton } \}$$ Now the set $P^*(P^*(P(U)))$ would act as a model for a theory under the second interpretation of your statement, and here this theory would be of the strength of second order arithmetic, which has enough strength to formalize MOST of traditional mathematics! 
A theory statisfiable in that model would be one with the following axioms: 
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x=y$
Atoms: $\forall x \forall y (x=\{y\} \to x=y)$
Multiplicity: $\forall x_1,..,x_n \exists y (singleton(y) \land y \neq x_1 \land ..\land y \neq x_n)$
Define: $elm(x) \equiv_{df} \exists y (x \in y)$
Comprehension: $\not \exists ! x (elm(x) \phi) \to \exists s \forall x (x \in s \leftrightarrow elm(x) \phi)$ 
Membership: $x \text { is a set of sets of singletons} \to elm(x)$
Now we define cardinal numbers as equivalence classes of sets of singletons under equivalence relation "disjoint bijection"
we say that "$x$ has disjoint bijection with $y$" if and only if  there exists a set $z$ such that $z$ is disjoint of both $x$ and $y$, and $x$ is bijective* (using non-ordered pairs) with $z$ and $z$ is bijective* with $y$. If that occurs we denote it by $|x|=|y|$. If $x$ had disjoint bijection with a subset of $y$, then we label that by $|x| \leq |y|$
Well-Foundedness: $\forall y \in x (y \text { is a cardinal number}) \land x \neq 0 \to \\\exists m \in x \forall y \in x (|m| \leq |y|)$
Of course naturals can be defined as the cardinal numbers that has every cardinal number smaller than or equal to them having an immediate predecessor cardinal number or otherwise being the first cardinal.
I should add that from the pure set theoretic point of view the above is not plausible, since skimming the world of sets of singletons of non singleton sets is not desirable at all. Possibly nicer approaches with this method is to have a theory with flat sets, i.e. only sets of quine atoms are there, i.e. no further iteration of the power set operator, and actually if we wont to stick to some mereological underpennings then we'd forbid having an empty set. This might be plausible philosophically. We can have a nearly unrestricted principle of comprehension over such structure, i.e. for any property $\phi$ holding of at least one Quine atom, then we can define the set of all Quine atoms satisfying that property. However, this turns to be weak also, at the strength of pure Mereology, which is insufficient to found mathematics. But it would also be interesting according to your qualification.
