By using Series you are assuming that the function is analytic at the point you are expanding around, which is false for your case: ultimately, you are expanding around $b=0$ for which $x[b]$ is divergent.
One way to get around this is to expand as follows:
Series[f[x[b]], {b, 0, 3}]
Then replace $x$ with its series form:
intermeadiateResult=
Series[f[x[b]], {b, 0, 3}]
/.x->Function[var,
Evaluate[
Normal[1/b Series[1/(1 - b), {b, 0, 3}]] /.b -> var+k
]
]
where $k$ is the regulator. It should be taken to $0$ at the very end; however, as it stands, we cannot take it to zero as the expression would diverge. At this point, the result looks like follows:
\begin{equation}
f\left(k^2+k+\frac{1}{k}+1\right)+b \left(-\frac{1}{k^2}+2 k+1\right) f'\left(k^2+k+\frac{1}{k}+1\right)+\frac{1}{2} b^2 \left(\left(-\frac{1}{k^2}+2
k+1\right)^2 f''\left(k^2+k+\frac{1}{k}+1\right)+\left(\frac{2}{k^3}+2\right) f'\left(k^2+k+\frac{1}{k}+1\right)\right)+\frac{1}{6} b^3
\left(\left(-\frac{1}{k^2}+2 k+1\right)^3 f^{(3)}\left(k^2+k+\frac{1}{k}+1\right)+3 \left(\frac{2}{k^3}+2\right) \left(-\frac{1}{k^2}+2 k+1\right)
f''\left(k^2+k+\frac{1}{k}+1\right)-\frac{6 f'\left(k^2+k+\frac{1}{k}+1\right)}{k^4}\right)+O\left(b^4\right)
\end{equation}
As an example, let us assume $$f(x)=\frac{1}{x}$$ In that case, we can get the result as follows:
Series[Normal[intermeadiateResult /. f -> Function[x, 1/x]], {k, 0,0}]
which yields $k$ independent result $(b - b^2)$.
One can check that it is not possible to get rid of $k\rightarrow 0$ limit for all $f$. For those functions, it simply means that the expression we are calculating is divergent so we should not have tried Taylor series to being with, but possibly a Laurent series.