For instance, if we take the set $M = \{f(x) = a_0 + a_1x + \cdots + a_nx^n\}$ and assume that if $f(x) = a_0 + a_1x + \cdots + a_nx^n$ with $0 < a_n$ then $0 < f(x)$ for large x, then this set is obviously bounded above by something like $h(x) = x^{n+1}$.
Knowing this, can this set still satisfy the Archimedean property if it isn't bounded by anything inside of $M$?
The way you defined it, your group can only have Archimedean property if $n=0$. If $n>0$, just take $1$ and $x$. Then for every natural number $m$, $m\cdot 1<x$, since eventually $x$ will be bigger than any natural number.