Example of a monoid required for union operation Consider $(G,\cup)$ to be a groupoid defined with respect to the union operation.
Let $G=\{A,B,C\}$ ,   where $C= A \cup B$ and $A$, $B$ are any arbitrary sets.

*

*$(G,\cup)$ is closed.


*Associative law holds for the group.


*Identity element seems to exists but doesn't as it isn't unique
$A \cup A=A=A \cup \{ \}$
Is it possible to define a semi group with respect to the operation mentioned above and obtain a monoid or even a group ?
 A: There are several issues in your question. In particular, it is ambiguous to call $G$ a groupoid, and later a group. To start with, $G$ is only a set, and union is not an operation on $G$, but on the set $P(G)$ of subsets of $G$. Moreover, it is not correct to say that $G$ is closed, for two reasons: first, you are actually not considering $G$ but $P(G)$. Moreover, it is not correct to say that $P(G)$ is closed. The right way would be to say that $P(G)$ is closed under the union operation (or simply under union).
Thus, I hope you will not mind if I first rephrase your question as follows:

Let $G$ be a set. Consider the set of all subsets of $G$, equipped
  with the union as operation. I already observed that this operation is
  associative. What type of algebraic structure does it define: a
  semigroup, a monoid, a group?

Answer. It defines a commutative monoid. You already observe that union
is associative and hence defines a structure of semigroup. It is a commutative semigroup since for all subsets $E$ and $F$ of $G$, $E \cup F = F \cup E$.
Furthermore, the empty set is the identity for this operation, since, for every subset $E$ of $G$, $E \cup \emptyset = \emptyset \cup E = E$.
The only case when you obtain a group is when $G$ is the empty set. Do you see why it is a group in this case? If $G$ is nonempty, the full subset $G$ has no inverse. Indeed, an inverse would be a subset $E$ of $G$ such that $G \cup E = \emptyset$, but this would imply $G = \emptyset$, a contradiction.
