Is it possible for a set to contain an element that does not have the defining property? Let $A:=\{x \ | \ \Phi(x)\}$ be a set and $a,b$ be objects such that $a=b$ and $b$ satisfies $\Phi(b)$ but $a$ does not satisfy $\Phi(a)$. Is it true that $a \in A$? A more concrete example would be $$B:=\{e^{2\pi ix} \ | \ x \in [0,1)\}$$ and $a:=1,b:=e^{2\pi i}$ (or any other integer multiple of $2\pi ix$). It then holds that $a \in B$ and thus by $b=a$ it should also hold that $b \in B$. However, is this conflicting with the concept of the defining property? I assume it is not because the set contains all the objects that have the value of any expression $e^{2 \pi i x}$ with $x \in [0,1)$, thus also containing $b$, however I am not sure since this might be some philosophical or set theoretic question that I am not familiar with and also touches on the question "what equality really is". Any comment is greatly appreciated.
 A: 
Let $A:=\{x \ | \ \Phi(x)\}$ be a set and $a,b$ be objects such that $a=b$ and $b$ satisfies $\Phi(b)$ but $a$ does not satisfy $\Phi(a)$.

That's impossible. By the subsitution principle, if $a = b$ and the proposition $\Phi(b)$ is true, then the proposition $\Phi(a)$ is true, too.

A more concrete example would be $$B:=\{e^{2\pi ix} \ | \ x \in [0,1)\}$$ and $a:=1,b:=e^{2\pi i}$ (or any other integer multiple of $2\pi ix$). It then holds that $a \in B$ and thus by $b=a$ it should also hold that $b \in B$.

That's right. In this example, $b = e^{2 \pi i}$, which means that $b = 1$, which means that $b = e^{2 \pi i \cdot 0}$, and so clearly $b \in B$.
In other words, the number $e^{2 \pi i}$ is a number of the form $e^{2\pi ix}$ where $x \in [0,1)$. The "obvious" choice of $x = 1$ fails to show this fact, but if we instead choose $x = 0$, we succeed in showing this fact.

However, is this conflicting with the concept of the defining property?

In my opinion, no, not at all. If we define a set $S = \{\text{Voltaire}\}$ whose sole element is the well-known 18th-century French author Voltaire, then we should not be surprised to learn that François-Marie Arouet is an element of $S$, because François-Marie Arouet is the same person as Voltaire.
A number can have multiple names, and it's still the same number no matter what name we call it by. And a number is either in the set or out of the set; it can't be in the set by one name, but out of the set by a different name.
A: $b=e^{2\pi i}$ and $c=e^{4\pi i}$ are both members of the set $\{ e^{\pi i x} : 0\le x<1\}$ because there is a value of $x$ for which $0\le x<1$ and $e^{2\pi i} = e^{4\pi i} =1,$ namely $x=0.$
If you want to express the set in the form $\{a: \Phi(a)\}$ then you can write $\big\{ z : \exists x\, \big( 0\le x<1 \text{ and } z = e^{\pi ix}\big) \big\}.$ Then $z=e^{4\pi i}$ is seen to satisfy the formula $\exists x\,\big(0\le x<1\text{ and } z = e^{\pi i x} \big),$ since $e^{\pi i 0}=e^{\pi i 4}.$
A: If your set consists of complex numbers then the elements are values and the two values are the same and the two expressions represent the same element of the set.
If your set consists of expressions $e^{2\pi i x}$ then $1$ is not an expression of that form and is therefore not an element of the set, and $e^{2\pi i }$ is not included either.
Motto: when you specify a set in this way, it is important to specify what kind of elements you are talking about. In the axioms of set theory this is made explicit. Quite often it can be implicit (we are talking about sets of real or complex numbers, and assume that all our sets are of that kind).
