Is there an analytical approximation that results in the following:
$\dfrac{x^{\theta-1}.(e^{-ax}-e^{-bx})}{\Gamma({\theta}).(a^{-\theta}-b^{-\theta})}\simeq \dfrac{x^{\theta^*-1}.e^{-cx}}{\Gamma({\theta^*}).c^{-\theta^*}}\ \ \ \ \forall x\in[0,\infty)\ \ ,\ \ b\geq a\ \ ,\ \ a,b,c,\theta,\theta^*>0$
Where the right hand side is the density of a gamma random variable with a shape $\theta^*$ and rate $c$.
I have already found some numerical examples for the approximation. For instance:
$\dfrac{x^{1-1}.(e^{-x}-e^{-2x})}{\Gamma({1}).(1^{-1}-2^{-1})}=2(e^{-x}-e^{-2x})\simeq \dfrac{x^{1.882-1}.e^{-1.256x}}{\Gamma(1.882).1.256^{-1.882}}$
Also, for the extreme cases an analytical approximation is easy to derive:
$\lim_{b\to a}\dfrac{x^{\theta-1}.(e^{-ax}-e^{-bx})}{\Gamma({\theta}).(a^{-\theta}-b^{-\theta})}= \dfrac{x^{(\theta+1)-1}.e^{-ax}}{\Gamma({\theta+1}).a^{-(\theta+1)}}$
$\lim_{b\to \infty}\dfrac{x^{\theta-1}.(e^{-ax}-e^{-bx})}{\Gamma({\theta}).(a^{-\theta}-b^{-\theta})}= \dfrac{x^{\theta-1}.e^{-ax}}{\Gamma({\theta}).a^{-\theta}}$
However, is there a general approximation that works for any arbitrary value of $b$?
An idea is to calculate a notion of distance between the true and the approximated distribution, which can then be optimized for the parameters in question. The simplest way to set this up would be to minimize the squared vertical distance between the two distributions
$$D(c,\theta^*)=\int_0^\infty(F(x;a,b,\theta)-F_\text{approx}(x;c,\theta^*))^2dx$$
It is clear from the form of the distributions and from the restrictions placed on the parameters, that the integrals converge for any choice of $a,b,c,\theta,\theta^*>0$. Also, it is easy to show that $D(c,\theta^*)$ is continuous and twice differentiable for all values of the parameters.
Defining the quantity
$$f(x;a,z):=\frac{a^z}{\Gamma(z)}x^{z-1}e^{-ax}$$
we can write the derivatives of $D$ with respect to $(c,\theta^*)$ as
$$\frac{\partial D}{\partial c}=-c\int_0^\infty f(x;c,\theta^*)\left(f(x;c,\theta^*)-\frac{a^{-\theta}}{a^{-\theta}-b^{-\theta}}f(x;a,\theta)+\frac{b^{-\theta}}{a^{-\theta}-b^{-\theta}}f(x;b,\theta)\right)dx$$
$$\frac{\partial D}{\partial \theta^*}=-\frac{\ln c-\psi(\theta^*)}{c}\frac{\partial D}{\partial c}+\\+\int_0^\infty\ln x~ f(x;c,\theta^*)\left(f(x;c,\theta^*)-\frac{a^{-\theta}}{a^{-\theta}-b^{-\theta}}f(x;a,\theta)+\frac{b^{-\theta}}{a^{-\theta}-b^{-\theta}}f(x;b,\theta)\right)dx$$
and by using the expressions for the integrals
$$\int_0^\infty f(x;a,w)f(x;b,z)dx=\frac{\Gamma(z+w-1)}{\Gamma(z)\Gamma(w)a^{-w}b^{-z}(a+b)^{w+z-1}}$$
$$\int_0^\infty\ln x ~ f(x;a,w)f(x;b,z)dx=\frac{\Gamma(z+w-1)}{\Gamma(z)\Gamma(w)a^{-w}b^{-z}(a+b)^{w+z-1}}\left(\psi(w+z-1)-\frac{w+z-1}{a+b}\right)$$
which results in the system of equations
$$\frac{\partial D}{\partial c}=-c\left[\frac{c\Gamma(2\theta^*-1)}{2^{2\theta^*-1}\Gamma^2(\theta^*)}+\frac{c^{\theta^*}\Gamma(\theta+\theta^*-1)}{(a^{-\theta}-b^{-\theta})\Gamma(\theta)\Gamma(\theta^*)}\left(\frac{1}{(a+c)^{\theta+\theta^*-1}}-\frac{1}{(b+c)^{\theta+\theta^*-1}}\right)\right]=0$$
$$\frac{\partial D}{\partial \theta^*}+\frac{\ln c-\psi(\theta^*)}{c}\frac{\partial D}{\partial c}=\\\left[\frac{c\Gamma(2\theta^*-1)}{2^{2\theta^*-1}\Gamma^2(\theta^*)}(\psi(2\theta^*-1)-\frac{2\theta^*-1}{2c})+\frac{c^{\theta^*}\Gamma(\theta+\theta^*-1)}{\left(a^{-\theta}-b^{-\theta}\right)\Gamma(\theta)\Gamma(\theta^*)}\left(\frac{\psi(\theta+\theta^*-1)-\frac{\theta+\theta^*-1}{a+c}}{(a+c)^{\theta+\theta^*-1}}-\frac{\psi(\theta+\theta^*-1)-\frac{\theta+\theta^*-1}{b+c}}{(b+c)^{\theta+\theta^*-1}}\right)\right]=0$$
Using the first equation, the second one may be written in the slightly more concise form
$$\frac{c\Gamma(2\theta^*-1)}{2^{2\theta^*-1}\Gamma^2(\theta^*)}\left(\psi(2\theta^*-1)-\psi(\theta+\theta^*-1)-\frac{2\theta^*-1}{2c}\right)+\frac{c^{\theta^*}\Gamma(\theta+\theta^*)}{\left(a^{-\theta}-b^{-\theta}\right)\Gamma(\theta)\Gamma(\theta^*)}\left(\frac{1}{(a+c)^{\theta+\theta^*}}-\frac{1}{(b+c)^{\theta+\theta^*}}\right)=0$$
Generally, this system would have to be solved numerically. One possible reduction of this problem that may yield a decent approximate solution is given by setting $\theta^*=\theta+1$, which matches the behavior of the two functions at the origin for $b\geq a$. We should expect that the optimal value for $\theta^*$ should satisfy this equation to some degree of accuracy. This reduction can be solved by solving the equation
$$c^{\theta}\left(\frac{1}{(b+c)^{2\theta}}-\frac{1}{(a+c)^{2\theta}}\right)=-\frac{a^{-\theta}-b^{-\theta}}{2^{2\theta}}$$
This always has a solution for the given parameter range, and this fact can be shown analytically.