= Solution
For a bounded real function on $[a,b]$, take a partition $a=t_0<\cdots<t_m=b$. Its lower and upper <Darboux sums> are
$$
L(f,P)=\sum_{j=1}^m\inf_{[t_{j-1},t_j]}f\,(t_j-t_{j-1}),\qquad U(f,P)=\sum_{j=1}^m\sup_{[t_{j-1},t_j]}f\,(t_j-t_{j-1}).
$$
The function is <Riemann integrable> when $\sup_P L(f,P)=\inf_P U(f,P)$; the common value is its <Riemann integral>. Equivalently, for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$. A <continuous function> is bounded and <Riemann integrable> on each compact interval, so all the following integrals exist.
Set $h(x)=x^{-1}\int_0^xf(t)dt$ for $x>0$. Strict decrease gives $f(x)<f(t)<f(0)$ for $0<t<x$. Integrating, with strict inequalities because the difference is positive on interior subintervals, gives
$$
f(x)<h(x)<f(0).
$$
The <intermediate value theorem> produces $g(x)\in(0,x)$ with $f(g(x))=h(x)$, and strict monotonicity makes this point unique. Thus \b[there is exactly one such $g(x)$ in $(0,x)$].
For the given <exponential function>, $h(x)=(1-e^{-x})/x$, so
$$
\boxed{g(x)=\log\left(\frac{x}{1-e^{-x}}\right).}
$$
The preceding strict bounds already prove $0<g(x)<x$; no limiting value at zero needs to be included in its domain.
For differentiability, a continuous strictly monotone function has a continuous inverse on its range. If it is differentiable at an interior point $t$ with $f'(t)\ne0$, the inverse is differentiable at $y=f(t)$, with derivative $1/f'(t)$. Indeed, setting $u=f^{-1}(y')$, inverse continuity gives $u\to t$ as $y'\to y$, and
$$
\frac{f^{-1}(y')-f^{-1}(y)}{y'-y}=\left(\frac{f(u)-f(t)}{u-t}\right)^{-1}\longrightarrow\frac1{f'(t)}.
$$
This one-dimensional inverse result needs no assumption that $f'$ is continuous. It is enough here that $f$ is differentiable and $f'<0$ at every positive argument.
The <fundamental theorem of calculus> and the quotient rule give
$$
h'(x)=\frac{xf(x)-\int_0^xf(t)dt}{x^2}=\frac{f(x)-h(x)}x<0.
$$
Now $g=f^{-1}\circ h$ and $g(x)>0$, so the inverse derivative formula applies at every relevant point. The <monotone mean-value point of an integral average> satisfies
$$
\boxed{g'(x)=\frac{h'(x)}{f'(g(x))}=\frac{f(x)-f(g(x))}{xf'(g(x))}>0.}
$$
Both numerator and denominator are negative. For the exponential example, this also agrees with $g'(x)=1/x-1/(e^x-1)>0$, since $e^x>1+x$ for $x>0$.
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