As a general remark, we may see the "Exponential to Ordinary" transformation of generating functions, $$ f(x):=\sum_{r=0}^\infty a_kx^k/k!\mapsto \tilde f(t):=\sum_{r=0}^\infty a_kt^k, $$  as an operator $\mathbb{C}[[x]]\to \mathbb{C}[[t]]$ . Since $n!t^n=\int_0^\infty (tx)^n e^{-x}dx$, the transformation can be computed analytically as $$\tilde f(t)=\int_0^\infty f(tx)e^{-x}dx,$$  at least for suitably convergent $f(x)$, and for special values of $t$. If the RHS is a convergent series $\sum_{r=0}^\infty b_kt^k$, and the equality holds,  for a set of values $t$ which accumulates within the disk of convergence, the identity of series $ \sum_{r=0}^\infty a_kt^k=\sum_{r=0}^\infty b_kt^k$ is then established  by the principle of isolated zeros. 

For instance, we may compute the transform of $f_r(x):=(e^x-1)^r$  for real negative values of $t$  in terms of the Euler's Beta function by a change of variable   in the integral:  
$$\tilde f_r(t)=\int_0^\infty (e^{tx}-1)^re^{-x}dx=(-1)^{r+1}t^{-1}\int_0^1(1-u)^n u^{-1/t-1}du=$$
$$=(-1)^{r+1}t^{-1}\frac{\Gamma(r+1)\Gamma(-1/t)}{\Gamma(r+1-1/t)}=\frac{r!}{(1-t)\dots(1-rt)}\ .$$
This computation gives your identity, since for the egf $f(x)$ of the $B_r$'s  we have $f(x)=e^{e^x-1}=\sum_{r=0}^\infty\frac{1}{r!}f_r(x)$ (in the sense of formal power series) so that  $\tilde f(t)=\sum_{r=0}^\infty \frac{t^n}{(1-t)\dots(1-rt)}\ .$

Incidentally, note that by an analogous computation you may, more generally, compute the ogf of the Stirling polynomials of the second kind, $B_r(z)$, starting by their egf $e^{z(e^x-1)}$.