Suppose $f$ is a strictly increasing, continuous function over $\mathbb{R}$, and $f(0) = 0$, $b > 0$. I need to show that for the change of variable $x = f(t)$,
$$\int_0^b f^{-1}(x)\,dx =\int_{0}^{f^{-1}(b)} t\,df(t) =bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,dt.$$ This looks like integration by parts, but we don't know how if $f$ is differentiable. Is the above true?
This question covers a lot of ground.
First, because $f$ is strictly increasing, the inverse $f^{-1}$ must be a strictly increasing continuous function. See here.
Next, there is a change-of-variables theorem for Riemann-Stieltjes integrals (Apostol Theorem 7.7) that justifies
$$\int_0^b f^{-1}(x)\,dx =\int_{0}^{f^{-1}(b)} t\,df(t)$$
Finally, since $f$ is increasing and $\alpha: t \mapsto t $ is continuous, $\alpha$ is Riemann-Stieltjes integrable with respect to $f$. By another well known theorem (Apostol Theorem 7.6), it follows that $f$ is integrable with respect to $\alpha$, and the integration by parts as you have written is valid:
$$\int_{0}^{f^{-1}(b)} t\,df(t) =bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,dt$$
No assumption that $f$ is differentiable is required.
Change-of-variables
Apostol's Theorem 7.7 states that
$$\int_{g(c)}^{g(d)} h(x)\, d\alpha(x) = \int_c^d g(h(t)) \, d \alpha(g(t)),$$
if $h$ is integrable with respect to $\alpha$ and $g$ is a strictly monotonic increasing continuous function.
Show how this applies to your question.