We observe that for any series $\omega(n) \notin \mathcal{O}(n)$ there is a series $\omega'(n) := \frac{\omega(n)}{n}$ with $\lim_{n \rightarrow \infty} \omega'(n) = \infty$. Therefore (using Chebyshev's inequality):
$$\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| > \omega(n)\sqrt{pq}) \leq \frac{Npq}{\omega(w)'n^2pq} \rightarrow 0 $$
as $n\rightarrow \infty$.
Since $$\forall\epsilon>0:\; \mathbb{P}_{n,p}(|e(G_{n,p}) - pN| > \omega(n)\sqrt{pq} \cap \mathbb{P}_{n,m}(Q_n) < 1 - \epsilon) \leq
\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| > \omega(n)\sqrt{pq})$$
this also holds for the intersection of those events.
We're now able to restrict our attention to the $m$ which are "linearly close" to the expected value $pN$, but we must consider every linear series $\omega(n)=xn$. So for every $\epsilon > 0$, as $n\rightarrow \infty$:
$$\mathbb{P}_{n,p}(\mathbb{P}_{n,m}(Q_n)<1-\epsilon)) \rightarrow 0 \\ \Leftrightarrow \forall x>0:\;\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| \leq xn\sqrt{pq} \cap \mathbb{P}_{n,m}(Q_n) < 1 - \epsilon)\rightarrow 0$$
Now we investigate the relation between $\mathbb{P}_{n,p}(Q_n)$ and $\mathbb{P}_{n,m}(Q_n)$:
$$\mathbb{P}_{n,p}(Q_n) = \mathbb{P}_{n,p}(\bigcup_{m\in\{1,\dots,N\}} (Q_n \cap e(G_{n,p}) = m))
\\= \sum_{m\in\{1,\dots,N\}} \mathbb{P}_{n,p}(e(G_{n,p}) = m)\cdot \mathbb{P}_{n,m}(Q_n)$$
For all $\epsilon$, we observe what happens, if more than $\epsilon$ values are at least $\epsilon$ far from $1$:
$$\mathbb{P}_{n,p}(\mathbb{P}_{n,m}(Q_n) < 1-\epsilon) > \epsilon \Rightarrow \mathbb{P}_{n,p}(Q_n) < 1 - \epsilon^2\\
\mathbb{P}_{n,p}(Q_n) > 1 - \epsilon^2 \Rightarrow \mathbb{P}_{n,p}(\mathbb{P}_{n,m}(Q_n) < 1-\epsilon) < \epsilon$$
Since we want the conditions to hold for every $\epsilon$ eventually, we can ignore the square:
$$\forall \epsilon \exists n_0\forall n > n_0:\;\mathbb{P}_{n,p}(\mathbb{P}_{n,m}(Q_n) < 1-\epsilon) > \epsilon \\
\Leftrightarrow \forall \epsilon \exists n_0\forall n > n_0:\;\mathbb{P}_{n,p}(Q_n) < 1 - \epsilon$$
We put together what we gained so far:
$$\mathbb{P}_{n,m}(Q_n) \rightarrow 1\\
\Leftrightarrow \forall x>0 \forall \epsilon,\epsilon'>0\exists n_0\forall n \geq n_0:\;\\
\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| \leq xn\sqrt{pq} \cap \mathbb{P}_{n,m}(Q_n) < 1 - \epsilon) < \epsilon'$$
The only thing that remains to do, is to show that we can express the latter condition without $\mathbb{P}_{n,p}$, i.e. just by counting elements, as stated in the lemma. We'll do this for each direction separately.
"$\Leftarrow$":
$\forall x>0 \forall \epsilon' > 0 \exists n_0\forall n \geq n_0:$
$$\max_{m\in\{m:|m-pN|\leq xn\sqrt{pq}\}} \mathbb{P}_{n,p}(e(G_{n,p})=m) = \mathbb{P}_{n,p}(e(G_{n,p})=pN) \leq \frac{1}{\sqrt{Npq2\pi}} + \epsilon'$$
$\Rightarrow\forall x>0 \forall \epsilon,\epsilon' > 0 \exists n_0\forall n \geq n_0:$
$$\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| \leq xn\sqrt{pq} \cap \mathbb{P}_{n,m}(Q_n) < 1 - \epsilon) \leq \#\{m:|m - pN| \leq xn\sqrt{pq}, \mathbb{P}_{n,m}(Q_n) < 1 -\epsilon \} \cdot (\frac{1}{\sqrt{Npq2\pi}} + \epsilon')\\
\leq\frac{\epsilon n\sqrt{pq}}{\sqrt{Npq2\pi}}+\epsilon'\epsilon n\sqrt{pq}
$$
Now observe that the set of the $m$ does not decrease as $\epsilon \rightarrow 0$, while both summands of the last line decrease. Therefore the only possible limit for both summands is $0$, which yields this direction of the statement.
"$\Rightarrow$":
$\forall x>0 \forall \epsilon > 0 \exists n_0\forall n \geq n_0:$
$$\min_{m\in\{m:|m-pN|\leq xn\sqrt{pq}\}} \mathbb{P}_{n,p}(e(G_{n,p})=m) = \mathbb{P}_{n,p}(e(G_{n,p})=pN-xn\sqrt{pq}) \geq \frac{1}{\sqrt{e^{x^2}Npq2\pi}} - \epsilon$$
We now assume the contrary of the right side:
$\Rightarrow\forall \epsilon' \exists x>0 \exists \epsilon > 0 \forall n_0\exists n \geq n_0:$
$$\mathbb{P}_{n,p}(|e(G_{n,p}) - pN| \leq xn\sqrt{pq} \cap \mathbb{P}_{n,m}(Q_n) < 1 - \epsilon) \geq \#\{m:|m - pN| \leq xn\sqrt{pq}, \mathbb{P}_{n,m}(Q_n) < 1 -\epsilon \} \cdot \frac{1}{\sqrt{e^{x^2}Npq2\pi}} - \epsilon'\\
\geq \frac{\epsilon n\sqrt{pq}}{\sqrt{e^{x^2}Npq2\pi}}-\epsilon'\epsilon n\sqrt{pq}$$
So convergence is impossible and this direction is proven indirectly.