* Let ${n \brace k}$ be a [Stirling number of the second kind][1].
* Let $A_n(x)$ be an [Eulerian polynomial][2]. Here
$$
A_n(x) = \sum_{i=0}^{n}i!{n \brace i}(x-1)^{n-i}.
$$
* Let $a_1(n,p,q)$ be the family of integer sequences such that
$$
a_1(n,p,q) = \sum\limits_{i=0}^{n} (p-q)^{n-i}{n \brace i}\prod\limits_{j=0}^{i-1}(qj+1).
$$
* Let $a_2(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$
B(x) = \exp\left(\sum\limits_{n=1}^{\infty}p^{n-1}A_{n-1}\left(\frac{q}{p}\right)\frac{x^n}{n!}\right).
$$
* Let
$$
R(n,m) = R(n-1,m+1) + p\sum\limits_{j=0}^{m}\binom{m+1}{j}q^{m-j}R(n-1,j), \\
R(0,m) = 1.
$$
* Let $a_3(n,p,q)$ be the family of integer sequences such that $a_3(n,p,q) = R(n-1, 0)$ for $n>0$ with $a_3(0,p,q)=1$.
* Let $a_4(n,p,q)$ be the family of integer sequences such that ordinary generating functions for it are $\frac{1}{G(0,x)}$ where $G(0,x)$ are continued fractions such that
$$
G(j,x)=1-\cfrac{(qj+1)x}{1-\cfrac{p(j+1)x}{G(j+1,x)}}.
$$
Note that
$$
G(0,x)=1-\cfrac{x}{1-\cfrac{px}{1-\cfrac{(q+1)x}{1-\cfrac{2px}{1-\cfrac{(2q+1)x}{1-\cfrac{3px}{1-\cfrac{(3q+1)x}{1-\cfrac{4px}{\ddots}}}}}}}}.
$$
* Let
$$
T(n,k,p,q) = (q(k-1)+1)T(n-1,k,p,q) + p(n-k+1)T(n-1,k-1,p,q), \\
T(n,1,p,q) = [n > 0], \\
T(n,0,p,q) = T(0,k,p,q) = 0.
$$
Here square bracket denotes [Iverson bracket][3].
* Let $a_5(n,p,q)$ be the family of integer sequences such that $a_5(n,p,q)=\sum\limits_{k=1}^{n}T(n,k,p,q)$ for $n>0$ with $a_5(0,p,q) = 1$.

I conjecture that
$$
a_1(n,p,q) = a_2(n,p,q) = a_3(n,p,q) = a_4(n,p,q) = a_5(n,p,q).
$$

Here is the PARI/GP program to check it numerically:

    a(n,p,q) = sum(i=0, n, (p-q)^(n-i)*stirling(n, i, 2)*prod(j=0, i-1, (q*j+1)))
    b(n,p,q) = p^n*sum(i=0, n, i!*stirling(n, i, 2)*(q/p-1)^(n-i))
    G(n,p,q) = my(CF = 1); for(j=0, n, CF = 1 - (q*(n-j)+1)*x/(1 - p*(n-j+1)*x/CF) + x*O(x^n)); CF
    F(n,p,q) = my(v1); v1 = Vec(1/G(n,p,q)); log(sum(i=0, n, v1[i+1]*x^i/i!) + x*O(x^n))
    upto1(n,p,q) = my(v1); v1 = vector(n, i, a(i, p, q))
    upto2(n,p,q) = Vec(1/G(n, p, q) - 1)
    upto3(n,p,q) = my(v1); v1 = vector(n, i, 0); v1[1] = 1; v2 = v1; for(i=2, n, v1 = vector(n, j, if(j==1, 1, (q*(j-1)+1)*v1[j] + p*(i-j+1)*v1[j-1])); v2[i] = vecsum(v1)); v2
    upto4(n,p,q) = my(v1); v1 = vector(n, i, b(i-1, p, q))
    upto5(n,p,q) = my(v1); v1 = Vec(F(n,p,q)); v1 = vector(n, i, v1[i]*i!)
    test1(n,p,q) = my(v1); v1 = upto1(n, p, q); v1 == upto2(n, p, q) && v1 == upto3(n, p, q)
    test2(n,p,q) = upto4(n,p,q) == upto5(n,p,q)

Is there a way to prove it?

  [1]: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
  [2]: https://oeis.org/wiki/Eulerian_polynomials
  [3]: https://en.wikipedia.org/wiki/Iverson_bracket