I am afraid that the proof given in Davies does not work in commutative case. The series given for $f^{-1}$: $$\sum_{n=0}^\infty (g-f)^{n}g^{-n-1}$$ indeed converges due to estimate on $||g^{-n}||$. But it has nothing to do with $f^{-1}$.It is apparently inspired by a geometric progression with $q=(g-f)g^{-n}$ (then $(1-q)^{-1}=((g-g+f)g^{-1})^{-1}=gf^{-1}$.) But in not-commutable case $q^{n}\neq (g-f)^{n}g^{-n}$ (there is no reson for $g-f$ or $f$ to commute with $g^{-1}$.)

It seems that if we assume $f\in C^1$ then we may require that both $||g-f||_{\infty}$ and $||g'-f'||_{\infty}$ are small. Hence $||g^{-1}||_{W}\le c(f)$. Consiquently $||q||_{W} < 1$ for sufficiently small $||g-f||_{C^{1}}$.

If $a_n$ are Fourier coefficients of $f$ it is enough to ask $\sum ||na_n||<\infty$ for $f \in C^1$. It would be great to avoid this extra condition.

Is there a counterexample for $f\in W \setminus C^1$?

I am afraid that the proof given in Davies does not work in not commutative case. The series given for $f^{-1}$: $$\sum_{n=0}^\infty (g-f)^{n}g^{-n-1}$$ indeed converges due to estimate on $||g^{-n}||$. But it has nothing to do with $f^{-1}$. It is apparently inspired by a geometric progression with $q=(g-f)g^{-1}$ (then $(1-q)^{-1}=((g-g+f)g^{-1})^{-1}=gf^{-1}$). But in not-commutable case $q^{n}\neq (g-f)^{n}g^{-n}$ (there is no reason for $g-f$ or $f$ to commute with $g^{-1}$).

It seems that if we assume $f\in C^1$ then we may require that both $||g-f||_{\infty}$ and $||g'-f'||_{\infty}$ are small. Hence $||g^{-1}||_{W}\le c(f)$. Consequently $||q||_{W} < 1$ for sufficiently small $||g-f||_{C^{1}}$.

If $a_n$ are Fourier coefficients of $f$ it is enough to ask $\sum ||na_n||<\infty$ for $f \in C^1$. It would be great to avoid this extra condition.

Is there a counterexample for $f\in W \setminus C^1$?