You've answered your own question. You say you're familiar with the fact that $(-b)c = -bc$ and $b(-d) = -bd$ in a ring. Now, if you write down a definition of multiplication that satisfies all of those properties, you get a ring structure on the set of equivalence classes, and $(-b)c = -bc$ and $b(-d) = -bd$ in a ring...!
Perhaps some extra generality will make this argument seem less circular. You have a rig $R$. This rig has an underlying additive monoid $M = (R, +)$. This abelian monoid has a Grothendieck group $G(M)$, which is the universal group into which $M$ maps. More precisely, there is a forgetful functor $F : \text{Ab} \to \text{CMon}$ from the category of abelian groups to the category of commutative monoids, and the Grothendieck group functor is its left adjoint. $G(M)$ consists of equivalence classes of formal expressions $m - n$ where $m, n \in M$ with addition defined as expected.
Now you want to put a ring structure on $G(M)$ compatible with the multiplication on $R$. It turns out that what you are actually constructing is the universal ring into which $R$ maps; more precisely, there is a forgetful functor from the category of rings to the category of rigs and you are constructing its left adjoint. Well, in any such ring we need
$$(a - b)(c - d) = ac - bc - ad + bd$$
as you well know, so this is the unique possible ring structure on $G(M)$ compatible with the multiplication on $R$ (and distributivity, inverses, etc.) if it is well-defined. This is a general property of adjoint functors (they are unique up to unique isomorphism if they exist). The only thing left is to verify that it actually works.
Another way to say the above is the following. Let $R$ be a rig and let $S$ be a ring, and let $\phi : R \to S$ be any rig homomorphism whatsoever. Then
$$(\phi(a) - \phi(b))(\phi(c) - \phi(d)) = \phi(ac) - \phi(bc) - \phi(ad) + \phi(bd).$$
There is nothing circular about this because $S$ is a ring and the above computation takes place in $S$.
I will try to write what I understood from Prof. Herbert Gross's teachings and innovation.
The addition operator operates on numbers, so a good knowledge of numbers can give us some insights. If we follow this PowerPoint by Prof. Gross, we have something very interesting to look at
In a similar way, $3$ inches is a $\color{grey}{quantity}$ in which the $\color{red}{adjective}$ is $\color{red}{3}$ and the $\color{blue}{noun}$ (unit) is $\color{blue}{inches}$ As quantities, $2$ fingers is not the same as $3$ inches. However, as $\color{red}{adjectives}$, the "$\color{red}{3}$" in "$3$ fingers" means the same thing as the $\color{red}{3}$ in "$3$ inches".
And from this presentation we have
If the students are asked to put two tiles, they will probably do this
And if they are asked to put those two tiles along with three more tiles, they will probably do this
$$ {\Huge \color{blue}{\blacksquare\blacksquare \qquad \blacksquare \blacksquare \blacksquare}} $$
And they will count those tiles as "one, two, three, four ,five", and thus know that there are five $\color{blue}{tiles}$.
$$
\begin{aligned}
{\Large \mathbf{\text{How to apply it in Algebra}}}
\end{aligned}
$$
Let's say we have $3~\color{red}{apples}$ and $2~\color{blue}{oranges}$ and if I ask you how many fruits do I have, you will answer $5$. So, what you have basically done is that you did the following translation
$$
3~\color{red}{apples} \rightarrow 3 ~\mathbf{fruits} \\
2 ~\color{blue}{oranges} \rightarrow 2~\mathbf{fruits}
$$
And then you added "3 fruits" and "2 fruits" in the same way as we added the tiles above.
Now, let's say we have any noun $x$, such that $3x$ means we have $3$ of those $x$(in the same way as $3$ fingers, or $3$ inches) and $2x$ means we have $2$ of those. If we say that we are given $3x$ and $2x$ then how many $x$ do I have in total? We can surely apply that tiles example once again to see things clearly, because it is stated that $x$ is any noun. Let's visualize $x$ by some strange looking figure and draw $3$ and $2$ of them
We again find, by counting, that we $5$ of $x$'s or $5x$.
As the question strictly doesn't allow Set Theory's or Peano's Axiom (works of Gottlob Frege, Bertrand Russell) definition of addition so I wouldn't touch on that and end my answer here.
Best Answer
As explained in the comments by several contributors, in most standard definitions of mainstream algebraic structures and applications, addition is supposed or defined to be associative (even in near-rings, the first operation is supposed to form a group). It is a useful convention to follow but one can define and study algebras with one, two or more internal binary operations (and even-ary operations) without this restriction but it is wise to use other names.
The two references you found are related to a line of research called "genetic algebra" by some, also "evolution algebra". See MSC 2010 subject 17D92, 17DXX started mainly in the forties by Etherington.
One of the main ideas is the combination/fusion/recombination of two gametes two make one new organism (that's close to an addition, but is non-associative since the order in which your ancestors reproduce matters) as well as the idea of seing familiar numbers as projection of trees. Etherington's name is now associated with various families of trees or equivalently to certain kinds of parenthesizing schemes.
See the Wikipedia entry for basic references on Genetic Algebra.
You might also be interested in other non-classical algebraic structures such as (planar) ternary rings, k-loops, operads but also, even if it is mainly associative, Tropical Mathematics.