The usual definition of a field is:
Definition. A set $F$ together with two binary operations $+$ and $\times$ is a field if and only if:
- $+$ is associative;
- $+$ is commutative;
- $+$ has a neutral element $0$;
- For every $a\in F$ there exists $b\in F$ such that $a+b=0$.
- $\times$ is associative;
- $\times$ is commutative;
- $\times$ distributes over $+$;
- There is an element $1\neq 0$ such that $a\times 1=a$ for all $a\in F$;
- For each $a\in F$, if $a\neq 0$ then there exists $x\in F$ such that $a\times x = 1$.
Now, you can certainly ask whether the axioms are independent. For example, it is easy to see that if you drop the assumption of commutativity for $+$, then you can deduce it from the other eight axioms; on the other hand, the real quaternions show that you cannot drop the assumption that $\times$ is commutative and deduce it from the other axioms.
So you could be asking whether Axiom 3 is independent of the other axioms. The main difficulty with dropping Axiom 3 is that without it Axiom 4 becomes unintelligible, and Axiom 9 is also problematic. So before we drop Axiom 3, we need to replace Axioms 4 and 9 with something else that, together with Axiom 3 give a field, but which make sense in the absence of Axiom 3.
There are plenty of ways of defining "abelian group" without specifying the existence of a neutral element (e.g., "for every $a,b\in F$ there exists $x\in F$ such that $a+x=b$") replacing 4 with something like this will automatically imply the existence of a $0$. Do you have something specific in mind?
(For a similar question, see for example this sci.math post by Dave Rusin where he discusses the independence of the axioms of a vector space, where he faces a similar problem with dropping "existence of $0$")
Added. While I was writing this, the title of the post was changed to "without an identity" instead of "without a $0$". You run into similar problems: if you drop Axiom 8, you need to replace Axiom 9 with something that still makes sense; depending on what you replace it with, it may or may not imply the existence of a multiplicative identity in the presence of the other axioms. Again, the question is whether you have something in mind or not.
If the question is just a poorly phrased way of asking if the one element ring is a field, the answer is that it is not considered to be a field. There are good reasons for this, even though the one element ring satisfies all the axioms except for the $1\neq 0$ clause of Axiom 8.