Edit, 7/4: Okay, I actually had a look through Bergstra, Hirschfield, and Tucker and I was making things too complicated. I think the cleanest way to set things up is to alter the definition of the weak inverse slightly (replacing axiom 2 by a different axiom), namely:
Revised definition: A weak inverse $y$ of an element $x$ in a monoid is an element satisfying $xyx = x$ and $yxy = y$.
This is the definition used to define inverse semigroups, and setting $x = 0$ in a ring immediately gives $y = 0$, so axiom 4 follows. Now we can define a weak field to be a commutative ring which has all weak inverses (with respect to multiplication), and we have:
Proposition: In a commutative monoid, weak inverses are unique.
Proof. Suppose $x$ has two weak inverses $y, z$. By definition this gives $x^2 y = x = x^2 z$ and $x y^2 = y, x z^2 = z$. In particular we have $x^2 y^2 = xy$ and $x^2 z^2 = xz$. We can now compute that
$$y = xy^2 = (x^2 z) y^2 = (x^2 y^2) z = xyz = (xz) y = (x^2 z^2) y = (x^2 y) z^2 = xz^2 = z. \Box$$
This feels like something someone should've told me a long time ago!
In any case, uniqueness makes things much easier; now if $A$ is a weak field we check as before that it's reduced and that every quotient $A/P$ is a weak field, we show that an integral domain is a weak field (this is now a property!) iff it's a field, so every quotient $A/P$ is a field, and we embed $A$ into $\prod A/P$ which is a product of fields and hence a weak field, and this is automatically a morphism of weak fields by uniqueness.
We also get a direct proof of axiom 3:
Corollary: In a commutative monoid, the weak inverse satisfies $(xy)^{-1} = x^{-1} y^{-1}$.
Proof. We just need to verify that
$$xy (x^{-1} y^{-1}) xy = (x^2 x^{-1})(y^2 y^{-1}) = xy$$
and
$$(x^{-1} y^{-1}) xy (x^{-1} y^{-1}) = (x^{-2} x)(y^{-2} y) = x^{-1} y^{-1}$$
and then it follows by uniqueness. $\Box$
Deprecated answer: Okay, it's actually pretty straightforward. The answer is yes. This is probably standard material to some people but I want to go through it step-by-step to be sure. For clarity, the entire problem here is that we don't know a priori that weak inverses are unique.
For starters, axiom 1 gives $a = a^2 a^{-1}$, and if $a$ is nilpotent then repeatedly applying this identity (to get $a = (a^2 a^{-1})^2 a^{-1}$ etc.) gives that $a = 0$, so axiom 1 implies that weak fields are reduced. If $A$ is a weak field this means it embeds into a product of fields, namely
$$\varphi : A \hookrightarrow \prod_P \text{Frac}(A/P)$$
where the product runs over all prime ideals $P$. To prove our desired result it suffices to show that $\varphi$ is a morphism of weak fields, where $\prod_P \text{Frac}(A/P)$ has the pointwise weak field structure, since this would show that every equational statement in the language of weak fields true in all fields is true in all weak fields.
First we need a slight generalization of the observation above that fields have a unique weak field structure.
Lemma 1: An integral domain $D$ has a weak field structure in at most one way, iff it is a field, in which case $a^{-1}$ is either the ordinary inverse if $a \neq 0$ or is equal to $0$.
Proof. If $a \neq 0$ then $a$ can be cancelled from axiom 1 which gives that $a^{-1}$ is the usual inverse; in particular the usual inverse must exist, so $D$ must be a field. Axioms 1 and 2 imply $0^{-1} = 0$. $\Box$
Lemma 2: If $A$ is a weak field and $P$ is a prime ideal of it (as a commutative ring), then the quotient map $\varphi_P : A \to A/P$ is a morphism of weak fields; in particular $A/P$ is a field and it has the unique weak field structure above.
Proof. Applying $\varphi_P$ to axiom 1 gives $\varphi(a)^2 \varphi(a^{-1}) = \varphi(a)$, hence that either $\varphi(a) = 0$ (so $a \in P$) or that $\varphi(a^{-1}) = \varphi(a)^{-1}$, where here this is the inverse in the usual sense (since $A/P$ is an integral domain). In particular $A/P$ is a field. This leaves open the possibility that we could have $\varphi(a) = 0$ but $\varphi(a^{-1}) \neq 0$. However, axiom 2 gives that if $\varphi(a^{-1}) \neq 0$ then $\varphi(a) = \varphi((a^{-1})^{-1}) = \varphi(a^{-1})^{-1} \neq 0$. So in fact $\varphi(a^{-1})$ is uniquely determined by $\varphi(a)$ via the unique weak field structure on $A/P$, and $\varphi_P$ is a morphism of weak fields. $\Box$
Corollary: The canonical embedding $\varphi$ of $A$ into a product of fields has the form $\varphi : A \hookrightarrow \prod_P A/P$, and is a morphism of weak fields if the RHS is given the pointwise weak field structure. In particular, weak inverses are unique in weak fields.
So in fact axioms 1 and 2 suffice to prove all equational statements in the language of weak fields which are true in all fields, and in particular they imply axiom 3. The uniqueness of weak inverses also implies that the weak fields are exactly the commutative von Neumann regular rings, and that the category of weak fields embeds fully faithfully into commutative rings.
This discussion appears to imply something about commutative vNr rings that I do not think is at all obvious otherwise, which is that the inclusion of commutative vNr rings into commutative rings has a left adjoint! If it exists it is given by formally adjoining weak inverses and I have no idea what it looks like concretely. Does that sound right to anyone else?
