Here is another proof, taken from "Markov Chains" by D. Revuz (lemma 1.5 page 190), always under the assumption that $(E, \mathscr{E})$ is countably generated.
The following answer is almost a paraphrase of the proof of Revuz.
By assumption, there is a sequence of finite partitions $\mathscr{P}_n$ of $E$, such that $\mathscr{P}_{n+1}$ is a refinement of $\mathscr{P}_n$, and $\mathscr{E}$ is generated by $\cup_{n \ge 0} \mathscr{P}_n$. For every $x \in E$, there exists a unique $E_n^x \in \mathscr{P}_n$ with $x \in E_n^x$.
Let $x \in E$ be fixed for the moment. Define $\lambda_x$ as the probability measure which is a multiple of $|Q_x|$. (if $Q_x = 0$, choose it as you want)
Then define a function $f_n$ on $E$ by $\displaystyle f_n(y) = \frac{Q_x(E_n^y)}{\lambda_x(E_n^y)}$ if $\lambda_x(E_n^y) > 0$ and $f_n(y) = 0$ otherwise.
By martingale convergence theorem, we have that $f_n$ converges $\lambda_x$-a.s. to the density of $Q_x$ with respect to $\lambda_x$. Hence $f_n^+$ converges $\lambda_x$-a.s. to the density of $Q_x^+$, and since $f_n$ are uniformly bounded, we have for all $A \in \mathscr{E}$ : $Q_x^+(A) = \lim_n \int_A f_n^+ \, d \lambda_x$
Now, if $A \in \mathscr{P}_k$, then for all $n > k$, $\int_A f_n^+ \, d \lambda_x = Q_{x,n}^+(A)$, where $Q_{x,n}^+$ is the positive part of the restriction of $Q_x$ to the $\sigma$-algebra generated by $\mathscr{P}_n$. It's easy to see that the map $x \to Q_{x,n}^+(A)$ is measurable, and so is the map $x \to Q_x^+(A)$.
We have hence proven that $x \to Q_x^+(A)$ is measurable for every $A \in \cup_{n \ge 0} \mathscr{P}_n$, and then a Dynkin class argument finishes the proof.