There is an intuitive approach to arithmetic of infinite ordinals:
"After all natural numbers comes the first infinite ordinal, $\omega$, and after that come $\omega + 1$, $\omega + 2$, $\omega + 3$, and so on ... After all of these come $\omega\cdot2$ (which is $\omega + \omega$), $\omega\cdot2+1$, $\omega\cdot2+2$, and so on, then $\omega\cdot3$, and then later on $\omega\cdot4$."
https://en.wikipedia.org/wiki/Ordinal_number#Ordinals_extend_the_natural_numbers.
However, before claiming we have an infinite set of multiplies $\omega\cdot n$ we need to make sure they are all unique.
Considering the Quadrants of a cyclically ordered group we can say that all natural numbers (excluding $0$) belong to the first quadrant of the group:
- $0 < \mathbb N < \frac{a}{2}$, whrere $a$ is the (imaginary) apex of the cyclic order of $\mathbb N$.
Applying the definition of $\omega$ we can say that $\frac{a}{2} = \omega$.
From this point of view:
- $\omega\cdot2 = 2\frac{a}{2} = a$;
- $\omega\cdot3 = 3\frac{a}{2} = -\frac{a}{2} = -\omega$;
- $\omega\cdot4 = 4\frac{a}{2} = 2a = 0$.
Thus, we only have four distinct multiples of $\omega$:
- $\omega\cdot n = \{0, \pm\omega, 2\omega\}$.
Would it be a valid statement?
If not, what is the main difficulty of the approach?
As explained in the comments, there is no group (cyclically ordered or otherwise) that contains 0, 1, and $\omega$ and whose operation is compatible with ordinal addition. Ordinals just don't fit into the framework of cyclically ordered groups.
I would recommend looking at the definition of the ordinals $\omega n$ for $n$ a natural number, and the definition of the linear order on ordinals. It should be easy to see that $\omega n < \omega (n+1)$ for all n: the latter contains the former as an element. In particular, this proves that the ordinals $\omega n$ are all distinct.