I solved the detailed balance equation subject to $\sum \pi_i=1$ and $\pi_i>0$ to get:
$$\sum_{i=0}^\infty \prod_{k=0}^{k=i-1}\frac{(A+kB)}{i!C}\pi_0 = 1$$
I'm having a tough time rearranging for $\pi_0$.
$$\sum_{i=0}^\infty \prod_{k=0}^{k=i-1}\frac{(A+kB)}{i!C}\pi_0 = 1$$ $$\sum_{i=0}^\infty \bigg(\frac{\pi_0}{\Gamma(i+1)C}\bigg)^i \prod_{k=0}^{k=i-1}(A+kB) = 1$$
Is it possible to do it?
On
Considering Jack D'Aurizio's answer, and replacing for clarity $$\frac{B\pi_0}C=x\qquad \text{and}\qquad \frac AB =a$$ we then need to look for the zero(s) of the function $$f(x)=\Gamma (a+2) \, _1F_1(a+2;3;x)-\frac{2}{a x^2}$$ $$f'(x)=\frac{1}{3} (a+2) \Gamma (a+2) \, _1F_1(a+3;4;x)+\frac{4}{a x^3}$$ So, the iterates of Newton method will be given by $$x_{n+1}=x_n \left(1+\frac{6-3\, a\, x_n^2\, \Gamma (a+2) \, \, _1F_1(a+2;3;x_n)}{12+a\, x_n^3 \,\Gamma (a+3) \,\, _1F_1(a+3;4;x_n)}\right)$$ which should not make any problem provided whe have an $x_0$ to start iterating. This is addressed in the edits of the post.
Since the problem as been reduced to a single parameter $a$, here are the results for a few values of the numerical solutions. $$\left( \begin{array}{cc} a & x \\ 0.1 & 2.03963 \\ 0.2 & 1.61757 \\ 0.3 & 1.38098 \\ 0.4 & 1.21673 \\ 0.5 & 1.09081 \\ 0.6 & 0.98855 \\ 0.7 & 0.90237 \\ 0.8 & 0.82783 \\ 0.9 & 0.76215 \\ 1.0 & 0.70347 \\ 1.1 & 0.65049 \\ 1.2 & 0.60228 \\ 1.3 & 0.55813 \\ 1.4 & 0.51750 \\ 1.5 & 0.47996 \\ 1.6 & 0.44519 \\ 1.7 & 0.41290 \\ 1.8 & 0.38286 \\ 1.9 & 0.35488 \\ 2.0 & 0.32881 \\ 2.1 & 0.30449 \\ 2.2 & 0.28179 \\ 2.3 & 0.26062 \\ 2.4 & 0.24087 \\ 2.5 & 0.22244 \\ 2.6 & 0.20527 \\ 2.7 & 0.18926 \\ 2.8 & 0.17436 \\ 2.9 & 0.16049 \\ 3.0 & 0.14759 \end{array} \right)$$
Edit
In order to start iterating Newton method, developing $f(x)$ as a very limited Taylor series around $x=0$, we have $$f(x)=-\frac{2}{a x^2}+\Gamma (a+2)+O\left(x\right)$$ which gives as a good starting point $$x_0=\sqrt{\frac{2}{a\, \Gamma (a+2)}}$$
To show how Newton method works, consider the case $a=0.1$. The iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 4.37168 \\ 1 & 3.25558 \\ 2 & 2.32495 \\ 3 & 2.03370 \\ 4 & 2.03963 \end{array} \right)$$ Now, for $a=3$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.166667 \\ 1 & 0.144357 \\ 2 & 0.147498 \\ 3 & 0.147592 \end{array} \right)$$
Working with Padé approximants, for $a>0.2$, a much better starting guess is $$x_0=6\frac{ 2a+4-\sqrt{6} \sqrt{-a^2-3 a+12 a \Gamma (a+2)-2}}{5 a^2+17 a-36 a \Gamma (a+2)+14}$$
Working the range $0.1\leq a \leq 10$, the following curve fit gives very good estimates $$x=\frac{1.2417}{a^{0.293448} \Gamma (a+2.54785)^{0.449602}}$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ A & 1.24170 & 0.02425 & \{1.19356,1.28984\} \\ B & 0.29345 & 0.00209 & \{0.28930,0.29759\} \\ C & 2.54785 & 0.04830 & \{2.45197,2.64373\} \\ D & 0.44960 & 0.00378 & \{0.44211,0.45710\} \\ \end{array}$$
For $a=0.1$, this would give $x_0=2.04435$; for $a=3$, this would give $x_0=0.146591$.
Your equation is equivalent to
$$ \sum_{i\geq 1}\frac{\Gamma\left(\frac{A}{B}+i\right)}{\Gamma(i+1)\Gamma(i-1) }\left(\frac{B\pi_0}{C}\right)^i = \frac{B}{A} $$ or to $$ \Gamma\left(\frac{A}{B}+2\right)\,\phantom{}_1 F_1\left(2+\frac{A}{B};3;\frac{B\pi_0}{C}\right)=\frac{2C^2}{AB\pi_0^2} $$ that is a trascendental equation involving a hypergeometric function. An approximate solution can be found by exploiting Gauss' continued fraction representation for $\phantom{}_1 F_1$, by exploiting Newton's method or through a combination of these approaches. I guess the right person to speak about this problem is Claude Leibovici, due to his experience in the field.