[step:Apply partial summation to pass from $\theta_{q,a}$ to $\pi_{q,a}$]
For $x>2$, the Stieltjes partial summation formula applied to the step function $\pi_{q,a}: [2,\infty)\to\mathbb N\cup\{0\}$ gives
\begin{align*}
\theta_{q,a}(x)=\pi_{q,a}(x)\log x-\int_2^x \frac{\pi_{q,a}(t)}{t}\,d\mathcal L^1(t).
\end{align*}
Equivalently,
\begin{align*}
\pi_{q,a}(x)=\frac{\theta_{q,a}(x)}{\log x}+\int_2^x \frac{\theta_{q,a}(t)}{t(\log t)^2}\,d\mathcal L^1(t).
\end{align*}
Because $\theta_{q,a}(t)\sim t/\varphi(q)$, for every $\varepsilon>0$ there exists $T\ge2$ such that for all $t\ge T$,
\begin{align*}
\left|\theta_{q,a}(t)-\frac{t}{\varphi(q)}\right|\le \varepsilon t.
\end{align*}
Splitting the integral over $[2,T]$ and $[T,x]$, the bounded initial segment contributes $O(1)$, while the tail gives
\begin{align*}
\int_T^x \frac{\theta_{q,a}(t)}{t(\log t)^2}\,d\mathcal L^1(t)
\sim \frac{1}{\varphi(q)}\int_T^x \frac{1}{(\log t)^2}\,d\mathcal L^1(t).
\end{align*}
Also $\theta_{q,a}(x)/\log x\sim x/(\varphi(q)\log x)$.
[/step]