Below I give a sketch of a proof.

First an approximation for the inverse incomplete gamma function, $Q^{-1}$, is needed.
Henceforth I assume that

$$
g(x)\sim \gamma x^s
$$

for large $x$ with $s>0$. Then ($0<b<1$)

$$
1-b^{1/g(x)}\sim (\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right),
$$

which is small for large $x$.

Now we use a formula due to Tricomi, which can be found here

$$
Q(\alpha+1,\alpha+\sqrt{2 \alpha} y) = \frac{1}{2} \text{erfc}(y),
$$

for $\alpha\rightarrow \infty$.

Setting $\alpha=x/2$ one gets for large $x$ (dropping constant and lower terms) the following approximation

$$
Q\left(\frac{x}{2},\frac{x}{2}\left(1+\frac{2 y}{\sqrt{x}}\right)\right) \sim \frac{1}{2} \text{erfc}(y) .
$$

Now we choose $y$ such that the right side is approximately equal to $(\gamma x^s)^{-1}\ln\left(\frac{1}{b}\right)$. With the help of the expansion of the inverse erfc, $\text{inverfc}$ (see e.g. here .)

$$
\text{inverfc} (y) \sim \frac{1}{\sqrt{2}} \sqrt{-\ln\left(\pi y^2 \ln \frac{1}{y}\right)} .
$$

Again using the fact that $x$ is large, we find $y\sim \sqrt{s} \sqrt{\ln x}$, keeping only the highest order in $\ln x$.

This gives

$$
Q^{-1}\left(\frac{x}{2},1-b^{1/g(x)} \right) \sim \frac{x}{2}\left(1+2\sqrt{ s}\sqrt{ \frac{\ln x}{x}}\right),
$$

independent of $\gamma$ and $b$.

Insertion and once more dropping terms of lower order in $x$ (assuming $f(x)/\sqrt{x}\rightarrow 0$ for $x\rightarrow \infty$)
$$
L(x) \sim Q\left(\frac{x}{2},\frac{x}{2}\left[1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right)\right]\right).
$$

Another asymptotic formula for the incomplete gamma functions comes into play (see, e.g., here):

$$
\Gamma(z,\lambda z) \sim (\lambda z)^z e^{-\lambda z}\frac{1}{z(\lambda - 1)},
$$

For this formula to be valid $\lambda > 1$.

Using the above and Stirling's formula for the $\Gamma(z)$ with $z=x/2$ and setting

$$
\lambda = \lambda(x) := 1+\frac{1}{\sqrt{x}} \left(2 \sqrt{s \ln x}- \frac{f(x)}{a}\right).
$$

after a lot of cancellations the result is

$$
L(x) \sim (2 \sqrt{s \ln x}- f(x)/a)^{-1}\rightarrow 0
$$

for $x \rightarrow \infty$.

Note that

$$
\lambda(x)^{x/2} \sim e^{x(\lambda(x)-1)/2},
$$

which cancels exactly with the exponential term in the approximation, thus leaving only the $\sqrt{x}/(\lambda(x)-1)$.

If we have $\lambda = 1$, i.e. $f(x) = 2 a \sqrt{s \ln x}$, another formula might be used (see, e.g. here)

$$
\Gamma(z,z) \sim \sqrt{\frac{\pi}{2}} z^{z-\frac{1}{2}}e^{-z} .
$$

With this I find

$$
L(x)\sim Q \left(\frac{x}{2},\frac{x}{2}\right)\sim \frac{1}{2},
$$

which is in disagreement with the numerical results. One has instead to take into account next order terms.