Is it possible to have a field without an additive identity? If I drop the axiom that Zero is the identity of an addition what consequences does this entail? What do I need to change to my axiomatization?

By definition it is not possible, but are there mathematical structures like a field but without an additive identity?

I don't have a concrete example/issue but I am interested in this question.
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 &quot;abelian group&quot; without specifying the existence of a neutral element (e.g., &quot;for every $a,b\in F$ there exists $x\in F$ such that $a+x=b$&quot;) 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 &quot;existence of $0$&quot;)
Added. While I was writing this, the title of the post was changed to &quot;without an identity&quot; instead of &quot;without a $0$&quot;. 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.
If I understand the definition of a field via abelian groups and distributivity as in Asaf's answer correctly, the perhaps surprising answer is no, you don't need to have an identity element in your axiom set for a field.  The trick lies in that you don't need an identity element in the axioms for a group.  Group theory, in fact, can get founded upon a single axiom, or several axioms which make no reference to a neutral element, if you really want that.  I've only glanced at things here, so see 1 2 and 3 for particular axiom sets.  Once you pick an appropriate axiom set for groups, you use the first part of Asaf's procedure for defining a field, which implies no mention of an identity element in the axiom set for fields.  The existence of inverses also doesn't mean one needs to talk about the identity element in the axiom set.  One could rewrite such existence statements via unary functions as one might do in the context of universal algebra, and this does capture the concept here, since inverses can get proved unique.