I am reading the book "Set Theory and the Continuum Problem" from Raymond and Melvin.
They have a a collection $V$, elements $x\in V$ are called sets, and $A \subseteq V$ is called class. The book starts with two axioms:
$P1\colon\, (\forall x)(x \in A \iff x \in B) \Rightarrow A =B $ and
$P2\colon\, (\forall A_1 \subseteq V)\dots (\forall A_n \subseteq V) (\exists B \subseteq V) (\forall x \in V) [x \in B \iff f(A_1,\dots,A_n,x)]$
where $f$ is a formula and $n=0$ is permitted.
The first theorem is
Theorem 1.1 Not every class is a set
This is proven by constructing the unique set $$ O = \{ x \in V \mid x \notin x\}$$ with $P1$ and $P2$ and then showing, that if $O$ would be set, one would get the Russell'S paradox $ O \in O \iff O \notin O$ and thus $O$ is a class that can't be a set.
However I wonder if there is actually a set with $x\in x$, because I can't imagine any. Or is $O = V$ ?
With just these two axioms, the answer is indeterminate. More to the point, $\sf NBG$ is more than just these two axioms.
Much like with $\sf ZF$, if you only include part of the axioms (or more specifically, if you omit Regularity) then you cannot decide the statement $\exists x(x\in x)$. It is consistent with the negation of Regularity that there is such $x$, and it is consistent that there are none (e.g., assuming Regularity).
The same is true here. If we only assume these two axioms, it is not nearly enough information to make a decision on whether or not some set is an element of itself.