So my thinking is that
$$P(A) = \{\{\emptyset\},\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\} $$
although it's clearly closed under union , assiociatve and the identity exists which is the empty set but still it's not a group since there are no inverses exists for any of the elements except for that of the empty set itself.
But is that valid to say , I mean how can I do it more rigorously?
The empty set is the identity but $B\cup C$ contains $B$ so if $B$ is non empty $B\cup C$ is non empty. Hence a nonempty element is not invertible.
Assume that inverses exist. Then:
$\{a\}\cup\{b,c\}=\{a,b,c\}=\{a,b\}\cup\{b,c\}\Rightarrow\{a\}=\{a,b\}$, a contradiction.
Your proof is valid as it stands.
It would be more compact to say that
$$\forall x \in P(\left\{a,b,c\right\} : a \in x\cup a $$ so the empty set cannot be the identity element, and then show that any other element can't be the identity either.
may be worth noting in this context that the power set of $X$ is a group under the operation of disjoint union, with the empty set as identity. if the complement of $A \subset X$ is $A'$ then the disjoint union of $A$ and $B$ is: $$ A \circ B = (A \cap B') \cup (A' \cap B) $$ proving associativity is a nice exercise, though it follows much more simply if the disjoint union is viewed as ordinary addition (mod 2) in $\mathbb{F}_2^X$
Let $~ X= \{1,2,3\}~$ and $~P(X)~$ be the power set of $~X~$. Examine if $~P(X)~$ is a group under the composition $~*~$ defined by $$~A*B= A\cup B ,~~~~~~~ \forall A,B \in P(X)~.$$
