This is a question in two parts.

Say that $\mathbf{On}$ is the proper class of all ordinal numbers in ZFC. We can define a binary operator over $\mathbf{On}$ which corresponds to the **commutative** version of ordinal addition; this has been called "Hessenberg addition" and "natural addition" before. It's also the operation you get by restriction of the $+$ operation from Conway's surreals to the subchain of ordinals (e.g. surreals with empty right set). I'll use the $+$ symbol for this operation over the ordinals.

$\langle\mathbf{On},+\rangle$ is a commutative monoid, which hence admits the notion of constructing a Grothendieck group $\mathrm{K}(\mathbf{On})$. The group $\langle\mathrm{K}(\mathbf{On}),+\rangle$ hence adds expressions such as $\omega$, $\omega-1$, $\omega^\omega - \omega^2 + 5$, etc. to the ordinals.

**Question 1:**is $\mathrm{K}(\mathbf{On})$ equivalent to Conway's "omnific integers" $\mathbf{Oz}$? In Conway's "On Numbers and Games," he defines an omnific integer $x$ as one which can be represented as a surreal number $\left \{ x-1 \mid x+1 \right \}$. Are these two classes isomorphic to one another?

It's also noteworthy that the field of fractions $Quot(\mathbf{Oz})$ is the full field $\mathbf{No}$ of surreal numbers. We can further turn $\mathrm{K}(\mathbf{On})$ into a ring $\langle\mathrm{R}(\mathbf{On}),+,\times\rangle$ by defining a new commutative operation called $\times$, called the "Hessenberg product", "Hausdorff product" or "natural product" of ordinals, which is commutative, associative, has an identity of 1, and distributes over the Conway normal form of the ordinal. A good definition for the Hessenberg product can be found on pages 24-25 of Ehrlich 2006.

**Question 2:**even if $\mathrm{K}(\mathbf{On})$ isn't isomorphic to $\mathbf{Oz}$, is $Quot(\mathrm{R}(\mathbf{On}))$ isomorphic to $\mathbf{No}$?

I'm tempted to answer in the negative for #1, as $\sqrt{\omega}$ is in $\mathbf{Oz}$, but is it in $\mathrm{R}(\mathbf{On})$? That is, given $\mathrm{K}(\mathbf{On})$ and ordinary commutative multiplication, is it the case that $\omega$ becomes a perfect square?

_{(Also, a last note - I'm aware that $\\mathbf{On}$ is a proper class. I'm not sure what foundational issues arise specifically in the above question, but I don't care how you want to handle them - NBG set theory, Grothendieck universes, whatever.)}