Has anything ever been done with the set $\{1,2,3,4,\ldots\}$ equipped with the operation $a \oplus b = a+b-1$ and the usual notion of multiplication? 
Definition. $$\mathbb{J} = \{1,2,3,\ldots\}.$$
We can refer to the elements of $\mathbb{J}$ as "joiners."

*

*The product of joiners is inherited from $\mathbb{Z}$.


*The sum of joiners will be defined by $$a \oplus b = a+b-1.$$

The motivation is that if we're thinking of the elements of $\mathbb{J}$ as non-empty finite totally ordered sets, then we can think of $a \oplus b$ as the result of "identifying" or joining the greatest element of $a$ to the least element of $b.$
This makes the $\mathbb{J}$ into a commutative monoid in two different ways. The good thing is that the identity elements of both these monoids are the same (namely $1$). The bad thing is that the distributivity law doesn't hold. Nonetheless, we do have the "variant distributivity" law:
$$(a(b \oplus c)) \oplus a = ab \oplus ac$$
Proof.
$$\mathrm{LHS} = a(b+c-1) \oplus a = (ab+ac-a) \oplus a = ab+ac-a+a-1 = ab+ac-1= \mathrm{RHS}$$
From this, we can deduce further "variant distributivity" identities, one for each natural number. For example, the $n=3$ case is $$a(b \oplus c \oplus d) \oplus a \oplus a = ab \oplus ac \oplus ad$$
This motivates the following definition.

Definition. Define that a flask is a pair of monoids on a common set, one denoted concatenatively, the other denoted with $\oplus$, such that:

*

*$\oplus$ is commutative

*their identity elements are equal and denoted $1$, and

*the variant distributivity law holds (on both sides).


It makes sense to speak of "commutative flasks" of course, the prototypical example being $\mathbb{J}$. There's lots of other flasks, though; for example, every ring $R$ can be viewed as a flask in the obvious way, and if $R$ is commutative, the corresponding flask will obviously be commutative. Note that we can view $\mathbb{J}$ as a subflask of $\mathbb{Z}$. This implies that subrings can have subflasks that aren't themselves subrings (because they don't contain $0$ and aren't closed under subtraction.)

Question. Has anything interesting ever been done with the algebraic structure I'm denoting $\mathbb{J}$ and/or the class of algebraic structures I'm referring to as flasks?
If so, is there a book or article that provides a good reference?

 A: I believe this could be related to an algebraic structure useful for studying set-theoretical solutions of the Yang-Baxter equation, the so-called braces. At least the funny distributive property appears in the theory of set-theoretical solutions of the Yang-Baxter equation.
Definition. An abelian group $(A,+)$ is a left brace if there is a multiplication $m\colon A\times A\to A$, $(a,b)\mapsto ab$, turning $A$ into a group such that 
$$a(b+c)+a=ab+ac$$ 
holds for all $a,b,c\in A$.
Similarly one defines right braces.
The connection to the Yang-Baxter equation is as follows. First, for each left brace $A$ 
one has a canonical solution $(A,r_A)$ of the Yang-Baxter equation: 
$$r_A\colon A\times A\to A\times A,\quad r_A(a,b)=(ab-a,(ab-a)^{-1}ab).$$
One also has the following.
Theorem. Let $X$ be a set, $$r\colon X\times X\to X\times X,\quad r(x,y)=(\sigma_x(y),\tau_y(x))$$ be a 
non-degenerate involutive solution of the Yang–Baxter equation. (Non-degenerate means that $\sigma_x$ and $\tau_x$ are permutations for each $x\in X$, and involutive means that $r^2=\mathrm{id}_{X\times X}$.) Then there exists a unique left brace structure over the group 
$$G=G(X,r)=\langle X:xy=\sigma_x(y)\tau_y(x)\rangle$$
such that its associated solution $r_G$ satisfies
$r_G(\iota\times \iota) = (\iota\times \iota)r$, where $\iota\colon X\to G(X,r)$ is the canonical map.
Furthermore, if $B$ is a left brace and $f\colon X\to B$ is a map such that $(f\times f)r=r_B(f\times f)$,
then there exists a unique brace homomorphism $\varphi\colon G(X,r)\to B$ such that $f = \varphi\iota$.
and $(\varphi\times\varphi)r_G = r_B(\varphi\times\varphi)$. 
The theorem is implicit in the work of Rump where braces were discovered:


*

*Rump, Wolfgang. Braces, radical rings, and the quantum Yang-Baxter equation. J. Algebra 307 (2007), no. 1, 153--170. MR2278047
You can find some interesting examples of braces here:


*

*Bachiller, David. Classification of braces of order $p^3$. J. Pure Appl. Algebra 219 (2015), no. 8, 3568--3603. MR3320237, arXiv: 1407.5224
See also this survey:


*

*Rump, Wolfgang. The brace of a classical group. Note Mat. 34 (2014), no. 1, 115--144. MR3291816
There is a non-commutative version of braces useful to study 
non-involutive solutions, see this paper.
Added. Rump proved that braces on both sides are in bijective correspondence with radical rings. The correspondence is as follows. If $R$ is a non-zero radical ring (for all $x\in R$ there is $y\in R$ such that $x+y+xy=0$) then $R$ with $a\circ b=ab+a+b$ is a two-sided brace. Conversely, if $A$ is a two-sided brace, then $A$ with $a*b=ab-a-b$ is a radical ring. 
