(Note I am the OP)

**Master equation**

The master equation can be written as: 
$$\frac{\mathrm d p_n{\left(t\right)}}{\mathrm dt}=\sum_{n'}\left\{W_{nn'}p_{n'}{\left(t\right)}-W_{n'n}p_n{\left(t\right)}\right\}$$
Where $p_n$ denotes the probability that the system is in state $n$ at time $t$. And $W_{nn'}\Delta t$ denotes the probability of the system changing from state $n'$ to the state $n$ in the time interval $\Delta t$.

**Specific case**

Let us say we have $n=1$ free enzyme, then:
$$ \frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=W_{01}p_1{\left(t\right)}$$
So what is $W_{01}$? This is the probability that the one enzyme is used up, let us call this $\mu$. 
$$\frac{\mathrm dp_0{\left(t\right)}}{\mathrm dt}=\mu p_1{\left(t\right)}$$
The average number of 'free enzymes' is given by: 
$$\left\langle n \right\rangle=p_1{\left(t\right)}$$
Differentiating this gives:
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=\frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}$$
But in our case,
$$ \frac{\mathrm dp_1{\left(t\right)}}{\mathrm dt}=-\mu p_1{\left(t\right)}$$
Thus
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu p_1{\left(t\right)}$$
$$\frac{\mathrm d\left\langle n \right\rangle}{\mathrm dt}=-\mu\left\langle n \right\rangle$$
From here with the assumption that $\mu\propto [\ce{S}]$ you can follow through my derivation on [Biology.SE][1]

The reason for the assumption $\mu\propto [\ce{S}]$ can be explained by the fact that $\mu$ is related to the rate of the reaction, which increases (approximately) linearly with substrate concentration.  If this assumption is not valid, please can you comment.

**Sources**

1.'Stochastic processes in physics and chemistry' by N.G. Van Kampen chapter V

2.  http://www.jpoffline.com/physics_docs/y4s7/advstatmech_ln.pdf

  [1]: https://biology.stackexchange.com/questions/38576/dwell-time-equations-for-atp-sythase?noredirect=1