I claim that
$$f_b(x) = 1 - \left(1 - 1/x^b\right)^{x^{b-1}} \le 2/x = g(x)$$
is true for every positive integer $b$ and for all $x \ge 1$. The claim appears to be true (even for non-integer $b$) when plotted, see: https://www.desmos.com/calculator/ysc7wdaati
How can we prove it, though?
I previously had asked a harder version of question, but got no answers. I think I might have made it much simpler this time, hopefully enough to warrant this new question.
Pull out the binomial series: $$(1+x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}x^k=1+\alpha x+\frac{\alpha(\alpha-1)}{2}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots$$ and you get $$\begin{align}f_b(x)&=1-\sum_{k=0}^\infty\binom{x^{b-1}}{k}(-1)^kx^{-bk}\\ &=x^{b-1}x^{-b}-\frac{x^{b-1}(x^{b-1}-1)}{2}x^{-2b}+\frac{x^{b-1}(x^{b-1}-1)(x^{b-1}-2)}{6}x^{-3b}-\cdots\\ &=x^{-1}-\frac{1}{2}x^{-2}\left(1-x^{-(b-1)}\right)+\frac{1}{6}x^{-3}\left(1-x^{-(b-1)}\right)\left(1-2x^{-(b-1)}\right)-\cdots\end{align}$$
You've noticed that $x\geq1$ is necessary, since that means all these negative powers of $x$ make things smaller.
So this is an alternating series with monotonic decreasing magnitudes, thus the error is the first term omitted when you approximate with the leading term.
You don't actually need the factor of $2$: $f_b(x)<1/x$ for all $x>1$, $b>1$ (equal if $x=1$ or $b=1$).
The plotting program you're using is running into severe numerical errors given the extreme powers you're throwing at it.