This is based on pp 240 - 242 of the 1965 edition Cox and Miller's "The Theory of Stochastic Processes". Please see M/G/1 queue: difference between $p_o$ and $p(0)$? for my earlier question.
We have:
$$0=p^{\prime}(x)-\lambda p(x) + \lambda p_0b(x) +\lambda\int_0^{\infty}p(x-y)b(y)dy$$
And
$$0 =-\lambda p_0 +p(0)$$
And if we take $\mathscr{L}\{p(x)\}$ to be $p^*(s)$ and $\mathscr{L}\{b(x)\}$ to be $b^*(s)$, we get: $$0=sp^*(s)-p(0)-\lambda p^*(s) + \lambda p_0b^*(s) + \lambda p^*(s)b^*(s)$$ And hence: $$p^*(s)=\frac{p(0)(1 -b^*(s))}{s-\lambda + \lambda b^*(s)}$$
I can follow this up to this point, but now lose the thread of the argument in two different, but I am sure related, ways...
Now $b(x)$ is the pdf of the service times of the arriving customers at the server (this is an M/G/1 queue) and Cox and Miller state "Let $\mu_b=-\frac{d}{ds}\{b^*(s)\}_{s=0}$ be the mean service time".
So $\frac{d}{ds}\{b^*(s)\} = \mathscr{L}\{-xb(x)\}$ (and I can see why, with $s=0$ that gives the mean) but I cannot see how to get an answer out of that.
Then Cox and Miller state for $s=0$:
$$p^*(0)=\frac{p(0)\mu_b}{1-\lambda\mu_b}$$
But if I take it on trust that $(1 -b^*(s)) = \mu_b:s=0$, I get: $$p^*(0)=\frac{p(0)\mu_b}{\lambda(1-\mu_b)}$$
I guess this is all tied up in the derivative of the Laplace transform, but someone point me in the right direction?
Noting that $b^{\ast}\left(0\right)=1$ (plug $0$ in for $s$ and remember that $b$ is a probability density and therefore integrates to $1$), we can rearrange the equation for $p^{\ast}\left(s\right)$ so that a few difference quotients appear, so when you take the limit $s\rightarrow 0$, you get the answer they give: $$\frac{p\left(0\right)\left(1-b^{\ast}\left(s\right)\right)}{s-\lambda+\lambda b^{\ast}\left(s\right)}=p\left(0\right)\frac{1-b^{\ast}\left(s\right)}{s}\cdot\frac{s}{s-\lambda\left(1-b^{\ast}\left(s\right)\right)}\textrm{.}$$ The difference quotient giving $\mu_{b}$ is $$\mu_{b}=-\left.\frac{d}{ds}b^{\ast}\left(s\right)\right|_{s=0}=-\lim_{s\rightarrow 0}\frac{b^{\ast}\left(s\right)-1}{s}$$ which appears in each of the quotients above, so the first quotient becomes $\mu_{b}$ and the second becomes $\frac{1}{1-\lambda\mu_{b}}$. This gives the answer they give.