Update: Eric Wofsey has demonstrated the conjecture in the commutative case below, and Tobias Kildetoft has provided a simple counterexample to the non-commutative claim.
This would-be replacement for the usual group axioms was suggested by the ##math IRC channel user Aleric, and no solution has been found so far. In its original form, the conjecture reads:
Let $(S,+)$ be a non-empty commutative semigroup satisfying the following "reversibility" axiom: for all $x,y \in S$, there exists a $z \in S$ such that $$x + y + z = x.$$ Then there exists an identity, i.e., an element $0 \in S$ such that $x + 0 = x$ for all $x \in S$.
Of course, if this is true then the identity must be unique, and $S$ becomes an Abelian group by applying the reversibility axiom to $0,x$.
I suppose one could just as easily drop the commutativity condition and formulate a stronger conjecture as follows:
Let $(S,\cdot)$ be a non-empty semigroup satisfying the following "reversibility" axiom: for all $x,y \in S$, there exist $z,w \in S$ such that $$xyz = wyx = x.$$ (Actually, it's not clear to me if we shouldn't ask for a sole element $z = w$ instead.)
Then there exists an identity, i.e., an element $1 \in S$ such that $x1 = 1x = x$ for all $x \in S$.
Again $S$ becomes a group by applying the reversibility axiom to $1,x$.
I have checked all the commutative semigroups of order $3$ and found none which satisfy this axiom and aren't groups, but this isn't very satisfactory. I am looking for a proof of either conjecture or a counterexample.