Inverse of integral transform $f(s)=\int_0^\infty g(x) \exp(-s g(x)) \mathbb{d}x$ Given $g(x)$ defined for positive reals, say $f(s)$ is defined as below
$$f(s)=\int_0^\infty g(x) \exp(-s g(x))  \mathbb{d}x.$$
Is there a relationship to named integral transforms, or a generic approach to obtain $g$ from $f$?
For instance, we can show the following holds through trial and error (notebook)
$$
\begin{align}
f(s) & = \frac{\sqrt{\pi }\ \text{erf}\left(\sqrt{s}\right)}{2 \sqrt{s}} &g(x)&=\frac{1}{(1+x)^2}\\
f(s)& = \frac{\Gamma \left(\frac{2}{3}\right)-\Gamma \left(\frac{2}{3},s\right)}{3 s^{2/3}} &g(x)&=\frac{1}{(x+1)^3}\\
f(s)& =\frac{-\cosh (s)+\sinh (s)+1}{s}  &g(x)&=\frac{1}{e^{x}}\\
f(s)&=\frac{e^{\left.\frac{1}{4}\right/s} \sqrt{\pi } (2 s+1)}{4 s^{5/2}} &g(x)&=\log ^2(x)\\
f(s) & = \frac{e^{-s} \left(-s+e^s-1\right)}{s^2} & g(x)&=\text{max}(0, 1-x)\\
f(s) & = \frac{1}{(s-1)^2} &g(x)&=\log(1+x)\\
f(s)&=\frac{e^{-s} (s+1)}{s^2} &g(x)&=1+x\\
\end{align}.$$
Edit Sep 7
Partial solution below works for increasing $g(x)=x+1$, but not for decreasing ones like $g(x)=(1+x)^{-2}$
Motivation: if loss after running $s$ steps of gradient descent steps on quadratic $Q$ decays as $f(s)$, then $i$th eigenvalue of Q quadratic decays as $g(i)$, hence knowing shape of loss curve over time would let us infer shape of quadratic $Q$ (background)
 A: Partial solution: Under certain conditions, this can be done using the inverse Laplace transform.
Let's restrict $g$ to be an increasing bijection $g: [0,\infty) \to [0,\infty)$, such that the integral defining $f$ converges. By integrating both sides from $s=t$ to $\infty$, we obtain
$$\begin{align}
\int_t^\infty f(s)ds &= \int_t^\infty \int_0^\infty g(x)e^{-sg(x)}dxds \\
&= \int_0^\infty \int_t^\infty g(x)e^{-sg(x)}dsdx \\
&= \int_0^\infty e^{-tg(x)}dx \\
&= \int_0^\infty (g^{-1})'(x) e^{-tx}dx \\
\end{align}$$
Thus, we have that $\int_t^\infty fds$ is the Laplace transform of $(g^{-1})'$. This means that $g^{-1}(y)$ is given by
$$\int_0^y \mathcal{L}^{-1}\bigg\{ \int_t^\infty f(s)ds\bigg\}(x) dx$$
from which the original $g$ can be recovered by inverting this expression.
A: The answer of Franklin Pezzuti Dyer is good, however, it involves one unnecessary assumption, namely that $g: [0,\infty)\rightarrow [0,\infty)$. This is not required. Let us do the substitution $y=g(x)$ in the original integral. This means $\mathrm{d}y=g'(x) \mathrm{d}x$ and $a=g(0)$, $b=g(\infty)$. Thus we have
$$f(s)=\int_{a}^{b}\mathrm{d}y\, \frac{1}{g'(x)}ye^{-s y}
=\int_{a}^{b}\mathrm{d}y\, \left(g^{-1}\right)^\prime(y)\, y\,e^{-s y},
$$
where $g^{-1}(y)=x$, i.e., is the inverse of $g(x)$. In other words $g^{-1}(g(x))=x$. The derivative of the inverse function $g^{-1}(y)$ is denoted as $\left(g^{-1}\right)^\prime(y)$. What remains to be done is to use the Heaviside function $\theta(y)$ to bring the last integral to the standard Laplace form. Let us assume that $0\le a < b$. We can write then
$$\int_{a}^{b} u(x)\mathrm{d}x = \int_{0}^{\infty} u(x)\left\{\theta(x-a)-\theta(x-b)\right\}\mathrm{d}x.$$
One should change a sign if $b>a$. With this in hands we can write
$$f(s)=\int_{0}^{\infty}\mathrm{d}y\,\left\{\theta(y-a)-\theta(y-b)\right\} \left(g^{-1}\right)^\prime(y)\, y\,e^{-s y}.$$
Assume now that the inverse Laplace transform of $f(s)$ exists $\mathcal{L}^{-1}[f].$ Then we have
$$\left\{\theta(y-a)-\theta(y-b)\right\} \left(g^{-1}\right)^\prime(y)\, y=\mathcal{L}^{-1}[f](y).$$
Of course, this is only a formal solution. It covers also the result of   Franklin Pezzuti Dyer as a partial case of $a=0$ and $b=\infty$ considering that $\mathcal{L}^{-1}[f]\in [0,\infty).$
Now, how the determination of $g(x)$ should be performed? As the first step one performs the inverse Laplace transform and looks at the codomain of $\mathcal{L}^{-1}[f]$. This yields the constants $a$ and $b$. Next one solves the differential equation for $g(x)$ with these boundary conditions.   Notice, that one interesting observation can be made about the table in OP. One can compute  $\mathcal{L}^{-1}[f]$ for many presented functions, they indeed feature $\theta$-functions consistent with the formal solution.
