Generalizing the natural numbers - has this been studied before?

The ordinals numbers generalize the natural numbers and satisfy a generalized induction principle. However, the algebraic properties of the ordinal numbers aren't so good. For example, ordinal sums are neither commutative, nor left-cancellative.

It was pointed out in the comments that these deficiencies can be remedied by working with the Hessenberg sum of ordinals. However, this introduces a new problem, which is that despite the fact that $1 \leq \omega$, nonetheless the equation $1 \boxplus x = \omega$ has no ordinal-valued solution (intuitively, it would have be $x = \omega - 1$). Here I am using $\boxplus$ to denote the Hessenberg sum.

So, I was trying to generalize the natural numbers in a different way, sacrificing proof by induction but preserving all those nice algebraic properties. I came up with the following idea, but haven't seen it anywhere.

Has the following idea been studied?

In what follows, lets take $\mathbb{N}$ to start at $1$. So

$$\mathbb{N} = \{1,2,3,\cdots\}.$$

Also, let $\mathbb{M}$ denote the positive rational numbers, with the convention $0 \notin \mathbb{M}.$

Then $\mathbb{N}$ embeds naturally into $\mathbb{Z}$ and $\mathbb{M},$ each of which embed naturally into $\mathbb{Q}$. We will try to recreate this system of embeddings.

Note that what follows is highly speculative, and there may be severe mistakes.

Let $\mathbb{N}'$ denote the set of all (non-zero) univariate integer polynomials with positive leading coefficient, ordered lexicographically. Then $\mathbb{N}'$ is closed under addition and multiplication, and these operations are commutative and associative on $\mathbb{N}'.$ We also have distributivity of multiplication over addition.

Furthermore, for all $a,b \in \mathbb{N}',$ I believe the following hold.

1. If $a < b,$ then for all $x \in \mathbb{N}'$ we have $x+a < x+b$ and $xa < xb.$

2. The following are equivalent. In particular, i implies ii and ii implies iii.

   i. There exists $x$ satisfying $x+a=b$.

   ii. $a < b$

   iii. There exists unique $x$ satisfying $x+a=b$.

3. For all $x \in \mathbb{N}'$ we have the following. If $x+a=x+b$, then $a=b.$ If $xa=xb$, then $a=b.$

Now let $\mathbb{Z}'$ denote the set of all univariate integer polynomials, ordered lexicographically. Thus $\mathbb{N}' = \{x \in \mathbb{Z}' \mid x > 0\}.$ Also, $\mathbb{Z}'$ is an integral domain.

Furthermore, for all $a,b \in \mathbb{Z}',$ I believe the following hold.

1. If $a < b,$ then for all $x \in \mathbb{Z}'$ we have $x+a < x+b.$

2. If $a < b,$ then for all $x \in \mathbb{N}'$ we have $xa < xb.$

3. There exists $x \in \mathbb{Z}'$ satisfying $x+a = 0.$

4. For all non-zero $x \in \mathbb{Z}'$ such that $xa=xb$, we have $a=b.$

Now recall that $\mathbb{M}$ is our notation for the positive rational numbers. By analogy, let $\mathbb{M}'$ denote the set of all equivalence classes of ordered pairs $(a,b) \in \mathbb{N}'$ with equivalence defined such that $$(a,b) \sim (a',b') \;\Leftrightarrow \; ab' = a'b.$$

Most likely, the set $\mathbb{M}'$ can be equipped with addition, multiplication in the usual way. Furthermore, define $$[(a,b)] < [(a',b')] \Leftrightarrow ab' < a'b.$$

I'm guessing that this is well-defined, and that the order relation interacts nicely with addition and multiplication.

Finally, we can construct an ordered field $\mathbb{Q}'$ in at least two different ways.

One way is to replace $\mathbb{N}'$ with $\mathbb{Z}'$ above. The order relation also needs to be defined in a more complicated way; presumably, the following definition will do the trick. $$[(a,b)] < [(a',b')] \Leftrightarrow (bb' > 0 \wedge ab' < a'b) \vee (bb' < 0 \wedge a'b < ab').$$

This is basically just the definition I got off Wikipedia's section on the formal construction of the rational numbers.

Another, most likely simpler approach would be to define $\mathbb{Q}'$ by gluing together two copies of $\mathbb{M}'$ with an element $0$ wedged between them. Presumably, both approaches to constructing $\mathbb{Q}'$ give us the same result, and this would be our first major theorem.

Finally, I'm guessing that we get the usual system of embeddings. So $\mathbb{N}'$ embeds naturally into $\mathbb{Z}'$ and $\mathbb{M}',$ and both these structures embed naturally into $\mathbb{Q}'.$ Indeed, its probably safe to view these embeddings as inclusions.

You may be interested in the surreal numbers, a totally ordered Field (capitalized because it is a proper class) in which the ordinal numbers with the Hessenberg operations embed. (Or just form the quotient Field of the ordinals with respect to the Hessenberg operations. The usual construction works because both operations are cancellative.)