I may be wrong here but does this not follow from the fact $S=\{f^n\:|\:n\geq 0\}$ is right Ore? Consider $f^{m} h$ for some $m\geq 0$ and $h\in R[x,\,x^{-1}]$. Let $y=x^{-1}$ and note that if $h\in R[x]$ then there is nothing to prove. For simplicity, suppose $h\in R[y]$ and has degree $k$, say. Then

$f^{m} h=f^{m} y^k h^\prime=y^k f^{m} h^\prime=y^k h^{\prime\prime}f^{l}$,

for $l\geq 0, h^\prime,h^{\prime\prime}\in R[x]$. Then $y^kh^{\prime\prime}\in R[x,x^{-1}]$ and you're done.

I may be wrong here but does this not follow from the fact $S=\{f^n\:|\:n\geq 0\}$ is right Ore? Consider $f^{m}, h$ for some $m\geq 0$ and $h\in R[x,\,x^{-1}]$. Let $y=x^{-1}$ and note that if $h\in R[x]$ then there is nothing to prove. For simplicity, suppose $h\in R[y]$ and has degree $k$, say. Then choose $h^\prime\in R[y]$ that is also of degree $k$. Then,

$f^{m} h^\prime=f^{m} y^k g=y^k f^{m} g=y^k g^{\prime}f^{l}$,

for $l\geq 0,\; g, g^\prime\in R[x]. Now, h=y^k \tilde{h}$ so we need to choose $g$ such that $g^\prime=\tilde{h}$. However, we can always do this because $S$ is right Ore. A similar argument exists if $h\in R[x,x^{-1}]$ (as opposed to just $R[y]$.