Does the set $S=\{x\mid x\in S\}$ exist? Consider the set $$S=\{x\mid x\in S\}.$$
For every element $x$, either $x\in S$ or $x\not\in S.$
If we know that $x$ is in fact element of $S$, then, by definition, $x\in S$ so it is true that $x\in S$.
If we know that $x\not\in S$, then, by definition, $x\not\in S$ so it is true that $x\not\in S$.
But if my question is what elements does $S$ contain, then the answer is "not certain."
 A: You haven't actually defined a set there. You've written down some symbols and are hoping that they define a set, but they don't.
The axiom schema which lets you define sets as "the things which satisfy some property" is the Axiom Schema of Comprehension: roughly, if $\phi$ is a predicate and $S$ is a set, then $\{x \in S: \phi(x)\}$ is a set. But you can't use that here because Comprehension selects a subset of a set which already exists; and $S$ isn't already known to be a set.
The axiom schema which lets you define sets by producing them as the image of another set is the Axiom Schema of Replacement: roughly, if $f$ is a function of classes and $S$ is a set, then $\{f(x): x \in S \}$ is a set. But you can't use this here with $f = \mathrm{id}$ because $S$ isn't known a priori to be a set.
A: This is an interesting question.
A set is determined by the elements it contains. That's the intuitive content of the assertion that two sets are equal if and only if they have the same elements.
So the set you're asking about is just $S$. It contains what it contains.
I think you are a little confused by the set builder notation $\{ \ldots | \ldots\}$.
That notation is usually invoked to construct a set of objects that satisfy some condition. For example
$$
E = \{ x | x = 2n \text{ for some integer } n\}
$$
creates the set of even integers by telling you what it means for an integer to be even.
In your example, the condition "$ x \in S$" does indeed tell you only what you already know by definition: the tautology "$x$ belongs to $S$ if and only if $x$ belongs to $S$".
Edit. @PatrickSevens answer says about the same thing as this one, but more carefully and more formally.
A: There is the concept in $\text{Set Theory}$ of a 'collectivizing relation', in other words, a relation that can be used to define sets.
Starting with any set $S$, 
$\tag 1 \text{There exists a set } X \text{ such that } X = \{x\mid x\in S \text{ AND } P(x) \}$
where P(x) is a statement about $x$ is collectivizing.
In particular, with
$\tag 2 P(x) \; :\; x = x$
we can create a new set
$\tag 3 A = \text{Set Defined by (1) \ (Set Builder Schema)}$
Exercise: Show that $A = S$.
The OP should have written,
Starting with a set $S$, what elements are in the set $S^{'} = \{x\mid x\in S\}$?
Ans: $S^{'} = S$.

In conclusion, you can only make sense out the expression
$\tag 4 S = \{x\mid x\in S\}$ in one way:
The expression (4) is a statement in set theory (perhaps an abuse of
  notation). The RHS builds a set, and the statement, $S = \text{RHS}$,
  is either TRUE or FALSE (and here (4) is always TRUE).
It is nonsensical to employ (4) to 'build a set $S$'.

A: To define an object, such as a set, means to provide a property that that object and only that object has. 
In your case, however, it turns out that every set $S$ has the property that $$S=\{x\mid x\in S\}.$$
This is just because every set is the set consisting of its elements. 
So you haven't successfully defined an object, you haven't picked out a particular set, since your property doesn't hold of one and only one object.
