Does anyone know of a non-trivial algebraic structure satisfying these four identities? Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?


*

*$(x+a)-a=x$

*$(x-a)+a=x$

*$(x+y)+a = (x+a)+(y+a)$

*$(x-y)+a = (x+a)-(y+a)$
Remark. The Abelian group of order $2$ doesn't satisfy the last two conditions.
Motivation. I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$
are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$

Below I will demonstrate that, if $+$ is associative, then $x+a = x$ and $x-a = x$ for all $x$ and $a$.  This is enough, I think, to qualify as a "trivial" algebraic structure, even though the underlying set can be as large as you like.
Beginning from (3):
\begin{align*}
(x+y)+a &= (x+a) + (y+a)\\
(x+y)+a &= ((x+a)+y) + a & \text{by associativity}\\
x+y &= (x+a)+y & \text{by (1)}\\
x &= x+a & \text{by (1)}
\end{align*}
Further, using (2) we find that $x = (x+a) - a = x - a$.
