Let $H(x) = \prod_{v=1}^n (a_v x+b_v) = \prod_{v=1}^n f_v(x)$. $\Lambda_n = (\lambda_1,\lambda_2,\ldots, \lambda_n)$ where $\Lambda_n$ is a sequence of non-negative integers. $|\Lambda_n| = \sum_{m=1}^n \lambda_m$. Denote $B_n^g(x) = e^{-g(x)}\frac{d^n}{dx^n}[e^{g(x)}]$, the expoenential complete bell polynomial. The partial is denoted as $B_{n,k}^g(x)$. S(n,k) is the stirling number of the first kind. ${r \choose \Lambda_n} = \frac{r!}{\lambda_1! \lambda_2! \cdots \lambda_n!}$ $$ B_{r}^{\ln H}(x) = (-1)^r \left(\sum_{k=1}^r (-1)^k S(r,k)\right) \left( \sum_{|\Lambda_n|=r}{r \choose \Lambda_n} \prod_{v=1}^n \left(\frac{a_v}{a_vx+b_v}\right)^{\lambda_v} \right) $$
Letting $Q_r(-1) = \sum_{k=1}^r (-1)^k S(r,k)$ we have an interesting way to define summations over a sequence of non-negative integers. $$ \sum_{|\Lambda_n|=r}{r \choose \Lambda_n} \prod_{v=1}^n \left(\frac{a_v}{a_vx+b_v}\right)^{\lambda_v} = (-1)^r \frac{B_r^{\ln H}(x)}{Q_r(-1)} $$ Since Bell Polynomials of this kind essentially taking r-th derivatives of a function $H(x)$, I like to think of this as an explicit definition of summations over $|\Lambda_n|=r$. But I would like to verify my work, this seems to be a very interesting result!
My proof:
I came across this due to my own work, which includes an identity I will be unable to provide a proof or reference to a proof, as my work is still incomplete and unpublished. But I have proven the following identity $$ B_{r,k}^H(x) = \sum_{j=k}^r \sum_{|\Lambda_n=j|} \rho(\Lambda_n,j-k) B_{r,j}^{\Lambda_n}(x) \prod_{v=1}^n f_v(x)^{j-\lambda_v} \tag{A} $$ The only thing important about the coefficient $\rho$ is that $\rho(\Lambda_n,0)=1$. Let $\Psi_n=(\psi_1,\psi_2, \ldots, \psi_n)$ $$ B_{r,k}^{\Lambda_n}(x) = \sum_{|\Psi_n|=r}{r \choose \Psi_n} \prod_{v=1}^n B_{\psi_v,\lambda_v}^{f_v}(x) \tag{A1} $$
In order to follow with this proof yourself it will be useful to check this reference in order to understand the derivation.
When using methods described in the reference we obtain $$ B_{r,k}^{f(g(x))}(x) = \sum_{m=k}^r B_{r,m}^g(x) B_{m,k}^f(g)\tag{B} $$ Letting $f(x) = \ln(x)$ it can be found using the reference $$ B_{n,k}^{\ln}(x) = (-1)^{n-k} S(n,k)x^{-n} \tag{C} $$ Before implementing (A) and (B), it will be easier to evaluate (A1). When we try to explicitly define $B_{\psi_v,\lambda_v}(x)$ we find an interesting case. Whenever $\psi_v=\lambda_v$ we find that $B_{\lambda_v,\Lambda_v}^{f_v} = a_v^{\lambda_v}$, otherwise the term is zero. Thus we set $\Psi_n = \Lambda_n$ and due to these sequences being equal, it also reasons that $r=j$ otherwise the terms will be zero. With this realization we can rewrite (A) $$ B_{r,r}^H(x) = \sum_{|\Lambda_n|=r} {r \choose \Lambda_n} \prod_{v=1}^n a_v^{\lambda_v} f_v^{r-\lambda_v}\tag{A*} $$ finally when we bring (A*), (B), and (c) and simplify we obtain $$ B_{r,k}^{\ln h}(x) = (-1)^{r-k}S(r,k) \sum_{|\Lambda_n|=r} {r \choose \Lambda_n} \prod_{v=1}^n \left( \frac{a_v}{a_vx+b_v} \right)^{\lambda_v} $$ Since this is the partial Bell Polynomial we take the sum $$ B_r^{\ln H}(x) = \sum_{k=1}^r(-1)^{r-k}S(r,k) \sum_{|\Lambda_n|=r} {r \choose \Lambda_n} \prod_{v=1}^n \left( \frac{a_v}{a_vx+b_v} \right)^{\lambda_v} $$ we can remove the portion that is independent of the $k$ variable and that is when we end up with $$ B_{r}^{\ln H}(x) = (-1)^r \left(\sum_{k=1}^r (-1)^k S(r,k)\right) \left( \sum_{|\Lambda_n|=r}{r \choose \Lambda_n} \prod_{v=1}^n \left(\frac{a_v}{a_vx+b_v}\right)^{\lambda_v} \right) $$