= Solution
If $\mathbb ES^2<\infty$, then $g''(1)=\lambda^2\mathbb ES^2$. Expand at $s=1$:
$$
g(s)=1+\rho(s-1)+\tfrac12g''(1)(s-1)^2+o((s-1)^2).
$$
Substitution into the previous quotient gives
$$
\Pi(s)=1+\left[\rho+\frac{g''(1)}{2(1-\rho)}\right](s-1)+o(s-1).
$$
Thus the stationary mean is
$$
\boxed{\mathbb E_\pi Q=\rho+\frac{\lambda^2\mathbb ES^2}{2(1-\rho)}.}
$$
Only a finite first service moment was stated. If the second moment is infinite and $\lambda>0$, the stationary mean is infinite; the formula has that extended-value interpretation. This follows as well by writing the second-order remainder as an integral of $g''(s)$ and taking $s\uparrow1$, where monotone convergence gives $g''(s)\uparrow\lambda^2\mathbb ES^2$. A finite <stationary distribution> need not have a finite mean.
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