*Simple Poisson Process.*

To gain a bit of intuition on why this result is true, it helps to consider a simple transition rate matrix:
$$
Q = \begin{bmatrix} -1 & 1 & & \\
& \ddots & \ddots & \\
& & &
\end{bmatrix}
$$ where all of the suppressed entries are equal to zero. This $Q$ is the transition rate matrix of a simple Poisson process with rate $1$.

In this case,
$$
m(\Gamma_{ij}) = \begin{cases}
1 & \text{if $j=i+1$} \\
0 & \text{otherwise}
\end{cases}
$$ and for any $t \ge s \ge 0$, integrating (1) over $[s,t]$ yields,
\begin{align*}
X_t - X_s &= \int_s^t \int_{\mathbb{R}} 1_{\Gamma_{X_{s-} X_{s-}+1}}(y) \nu(dt,dy) \\
&= \nu((s,t),(0,1))
\end{align*}
By definition of a Poisson random measure, the random variable $\nu((s,t),(0,1))$ has a Poisson distribution with rate $t-s$. Since $X_t$ also has independent increments, the process $\{ X_t \}$ started at the origin is a simple Poisson process with rate $1$, as we expected.

*Proof of (1).*

We will take for granted that the SDE (1) is well-posed, in the sense there exists a pathwise unique solution to (1) over any time interval.

Set $t_0=0$ and let $i_0 \in M$. The continuous time Markov chain $\{ X_t \}$ can be represented by a sequence of random jump times $\{ t_k \}$ and a Markov chain $\{ i_k \}$ called the embedded chain associated to $\{ X_t \}$. In particular, for any $t \ge 0$,
$$
X_t = i_k \quad \text{if $t_k \le t < t_{k+1}$}
$$
Moreover, the distributions of the jump times and embedded chain are given by $$
\mathbb{P}( t_{k+1} -t_k \mid X_{t_k} = i ) = \operatorname{Exp}(q_{i}) \;, \quad \text{and} \quad
\mathbb{P}( i_{k+1}=j \mid X_{t_k} = i) = \frac{q_{ij}}{q_i} \;.
$$
This representation is quite standard and shows that the process $\{X_t\}$ is a càdlàg Markov jump process. The proof given below shows that this representation is equivalent to (1) in a weak or distributional sense. BTW, this representation is used in the Doob-Gillespie algorithm for simulating continuous-time Markov chains.

Back to the proof, integrate (1) over the interval $[0,t]$ to obtain,
\begin{align*}
X_t - X_0 &= \sum_{i,j \in M} \int_0^t \int_{\mathbb{R}} (j-i) 1_{\Gamma_{ij}}(y) 1_{\{X_{s-}=i\}} \nu(ds,dy) \\
&= \int_0^t \sum_{i,j \in M} (j-i) 1_{\{ X_{s-}=i \}} \nu(ds, (0,q_{ij}) ) \\
&= \sum_{\substack{0 \le k \le N(t) \\ j \in M}} \int_{t_k}^{t_{k+1}} (j-i_k) \nu(ds, (0,q_{i_k j} )
\end{align*}
where we have introduced the sequence of stopping times
$$
t_{k+1} = \inf\left\{ t>t_k : \sum_{j \in M} \int_{t_k}^t \nu(ds, (0,q_{i_k j})) = 1 \right\}
$$ with $t_0=0$ and the Markov chain $\{ i_k \}$ for $0 \le k \le N(t)$ where $N(t)$ is the total number of jumps that occur in $X_t$ over $[0,t]$. Define $t_{k,j} = \inf\{ t>t_k : \int_{t_k}^t \nu(ds, (0,q_{i_k j})) \}$ and set $\delta t_{k,j} = t_{k,j}-t_k$.

Conditional on $X_{t_k}=i_k=i$, the random variables $\delta t_{k,j}$ are mutually independent exponential random variables with $\mathbb{P}(\delta t_{k,j} \mid i_k=i) = \operatorname{Exp}(q_{ij})$ and $t_{k+1} - t_k = \min_{j \in M} \{ \delta t_{k,j} \}$, it directly follows that
\begin{align*}
\mathbb{P}(t_{k+1}-t_k \mid i_k=i) = \operatorname{Exp}( q_i) \quad \text{and} \quad
\mathbb{P}(i_{k+1} = j \mid i_k=i) = \frac{q_{ij}}{q_i}
\end{align*}
which shows that the transition rate matrix of $X_t$ satisfying (1) is $Q$.

*Application of (1)*

While more involved than the simpler representation given in the proof, the representation (1) is particularly useful in approximation methods for continuous-time Markov chains like tau-leaping. For more info about this application, check out