So $$(F,+,*) $$is a field iff $$(F,+)\quad \text{and}\quad (F\setminus \{0\},*)$$ are abelian goups. $F\setminus \{0\}$ denotes F without the member zero.
So $$(F,+)$$ is sure a commutative group :
Identity exists which is $0$
Associativity exists
Commutative.
Inverse exists such that $-a+a=0$
is closed under multiplication.
But when we come to $(F\setminus \{0\},*)$
I couldnt find the inverse for all $a$ which is a member of $F$ Find me a number $b$ such that $a*b=0$ other than $0$?
So how come $(F\setminus \{0\},*)$ is a commutative/abelian group?
You're conflating two different operations. Note that the group operation in $(F\setminus \{0\}, *)$ is multiplication, not addition! You are correct that the reals without zero are not an abelian group under addition, but they are an abelian group under multiplication.
Put another way: when you go looking for inverses in $(F\setminus \{0\}, *)$, you're trying to multiply to $1$, not $0$. For example, in the reals without zero, the inverse of $3$ is ${1\over 3}$ - the "identity" element is $1$, and $3\cdot {1\over 3}=1$.
Note that this means that the multiplicative group of a field is not a subgroup of the additive group of the field!