Asymptotic solution for a first order ODE Simplified question*:
Given $f(t)$ that satisfies $f'(t)>0$,  $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$ we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of
$$ \dot{H}=F $$
Can we approximate H by F?
Specifically, I want to show that
$$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s+C)}=1 $$
for some constant C. $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.
If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.
* thanks to Willie Wong helpful suggestion.

The original question (to make the motivation more clear):
I have the following differential equation:
$$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$,  $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.
I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.
The motivation behind this solution is
$$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$
since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve
$$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$
for some constatnt C.
Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.
 A: The simplest cases are the linear cases where $f(t) = \alpha t$, and $H = \alpha^{-1}F$. Modelling on those cases, I will show that if asymptotically $f'(t)$ is bounded below (so if $f$ is asymptotically superlinear) than the desired conclusion hold. 

The first step is to prove that the conclusion holds if $H$ is eventually $< F$, using merely a convexity argument. 
Suppose there exists some $t_0$ such that for every $t > t_0$ we have that $H(t) < F(t)$. Notice that $\dot{H} = F$ and $\ddot{H} = \dot{f} F > 0$ we have that $H$ is convex. We are interested in the quantity 
$$ H^{-1}(s) - F^{-1}(s) $$
which is positive for all sufficiently large $s$ when $H$ is eventually $< F$. 
By convexity, we have that for any $r < s$
$$ H^{-1}(s) < H^{-1}(r) + (H^{-1})'(r) \cdot (s-r) $$
Since $H < F$ we can shoose $r = H\circ F^{-1}(s)$ and get
$$ H^{-1}(s) < F^{-1}(s) + (H^{-1})'\circ H \circ F^{-1}(s) \cdot (s - H\circ F^{-1}(s)) $$
Now
$$ (H^{-1})' = \frac{1}{H'\circ H^{-1}} $$
so we get
$$ H^{-1}(s) < F^{-1}(s) + \frac{1}{s} \cdot (s - H\circ F^{-1}(s)) $$
So in particular, you have that $H^{-1}(s) - F^{-1}(s)$ is bounded, since $H\circ F^{-1}$ is increasing. 
Using that $F^{-1}$ grows unboundedly this implies
$$ H^{-1} - F^{-1} = o(F^{-1})$$
as desired. 

Notice that in this argument I have not explicitly used the conditions $f' = \omega(t^{-1})$ and $\log(f') = o(f)$; however, the condition that $H(t) < F(t)$ can be derived from strengthened versions of your assumption. 
For example, a sufficient condition to guarantee that $H(t) < F(t)$ eventually for every solution $H$ is that $f'(t) > 1 + \epsilon$ for all sufficiently large $t$. 

This can be generalized further. Suppose that $f'(t)$ is lower-bounded by $\delta > 0$.
Define the function $\tilde{f}(\tau) = f(2 \delta^{-1} \tau)$. Define $\tilde{F}$ and $\tilde{H}$ analogously. We have that $\tilde{F} = F(2 \delta^{-1}\tau)$ and $\tilde{H} = \frac{\delta}{2} H(2\delta^{-1}\tau)$
Applying the above argument we get
$$ H^{-1}(2 \delta^{-1} s) - F^{-1}(s) < \frac{2}{\delta} - \frac{1}{s} H\circ F^{-1}(s) $$
Now
$$ F^{-1}(s) = f^{-1} \ln(s) $$
so 
$$ F^{-1}(2 \delta^{-1} s) = f^{-1} ( \ln(s) + \ln 2 - \ln \delta) $$
Using that $f'$ is bounded below by $\delta$, we have that
$$ |F^{-1}(2 \delta^{-1} s) - F^{-1}(s) | < \delta^{-1} \ln(2 \delta^{-1}) $$
is also bounded, so that we conclude 
$$ H^{-1}(s) - F^{-1}(s)$$
is bounded, and the unbounded growth of $F^{-1}$ takes care of the rest. 

The remaining case is when $\liminf_{t\to\infty} f'(t) = 0$, which is probably the case you are really interested in to boot, but unfortunately I don't yet see a proof for. 
