Here is a proof that $a_1(n, p, q) = a_2(n,p,q)$.
The generating function for the Stirling numbers of the second kind is $$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$ Also $$\prod_{j=0}^{i-1}(qj+1) = q^i i!\,\binom{1/q+i-1}{i}.$$ So \begin{align*} \sum_{n=0}^\infty a_1(n,p,q) \frac{x^n}{n!} &=\sum_{i=0}^\infty \left( q\over p-q\right)^i (e^{x(p-q)} -1)^i\binom{1/q +i-1}{i}\\ &=\left( 1-\displaystyle\frac{q}{p-q} (e^{x(p-q)}-1)\right)^{-1/q}\\ &=\left(\frac{p-q}{p-qe^{x(p-q)}}\right)^{1/q}. \end{align*}
The generating function for the Eulerian polynomials is $$ \sum_{n=0}^\infty A_n(t) \frac{x^n}{n!} = \frac{1-t}{e^{(t-1)x-t}}. $$ Integrating with respect to $x$ gives \begin{align*} \sum_{n=1}^\infty A_{n-1}(t) \frac{x^n}{n!} &= \frac{1}{t}\left[ \log\left(1-t\over e^{(t-1)x} -t\right) +(t-1)x\right]\\ &=\frac{1}{t}\log\left( 1-t \over 1-te^{(1-t)x}\right). \end{align*} Thus \begin{align*} \log B(x) &= \sum_{n=1}^\infty p^{n-1}A_{n-1}\left(q\over p\right)\frac{x^n}{n!}\\ &=\frac{1}{q}\log\left( 1-q/p\over 1-(q/p) e^{(1-q/p)px} \right)\\ &=\frac{1}{q}\log\left(p-q\over p-q e^{(p-q)x} \right), \end{align*} so $$B(x) = \left(\frac{p-q}{p-qe^{x(p-q)}}\right)^{1/q} =\sum_{n=0}^\infty a_1(n,p,q) \frac{x^n}{n!}.$$