Multiplicative identity being equal to additive identity in a field Is it even possible? What consequences would this have if it is possible?

My attempt:
Let us call this hypothetical universal identity $e$.

Fields require distributivity, right?
$(a-a)a^{-1} = aa^{-1} - aa^{-1} = e-e = e$

But calculating without using distributivity $(a-a)a^{-1} = ea^{-1} = a^{-1}$

So any multiplicative inverse must be the identity. Then we can not have elements other than $e$ regardless of how we try and define addition?
When $0$ is the additive identity for a ring, for any element $x$ in the ring, $x*0=x*(0+0)=x*0+x*0$. Subtracting $x*0$ from both sides of $x*0=x*0+x*0$ tells you that $x*0=0$.

If now $x$ is posited to be the multiplicative identity, it says that $x=x*0=0$. So such a ring is necessarily $\{0\}$.

As people have mentioned in the comments, the zero ring is excluded as a field by standard field axioms.
(1) Let $id_{+}$ = 0 = $id_{\times}$ in distributive system $\langle S, +, \times \rangle$.
By distributivity, $(0)\times A = (0+0)\times A$, so $A = A+A$ for all $A$ in $S$.

(2) Also, given $A+A = A$ for all $A$ in a distributive system 
with $id_{+}$, $id_{\times}$, and $-id_{\times}$ exist, then $id_+ = 0 = id_{\times}$.

(3) Given $id_{+}$ = 0 = $id_{\times}$ without property $A+A = A$ for all $A$ in $S$,
Implies that $\langle S, +, \times \rangle$ is a non-distributive system. Example:
Let $S = \mathbb R$, with A\times B := AB + A+B$. $id_\times = 0$, $A+A = A$ only for $A=0$,
But $\times$ does not distribute over real $+$.