Yes , the series $\beta_k(x):=\sum_{n\ge0} b(n,k) x^n$ are indeed rational functions, with the poles as you said.
One step back. The Exponential Generating Function of the polynomials $\big\{\sum_{k\ge0} b(n,k)t^k \big\}_{n\ge0}$ is $$\sum_{n\ge0}\Big(\sum_{k\ge0} b(n,k) t^k \Big)\frac{x^n}{n!} =e^{t(e^x-1-x)}$$ by the combinatorial meaning of the exponential of an EGF. Hence also, for any $k\ge0$, $$ \sum_{n\ge0} b(n,k) \frac{x^n}{n!}=\frac{(e^x-1-x)^k}{k!}.$$
To convert the latter into an ordinary Generating Function, apply the usual transformation $u(x)\mapsto \frac{1}{ x}\int_0^{\infty}u(s) e^{-\frac{s}{x}}ds$
$$\sum_{n\ge0} b(n,k) x^n =\frac{1}{k!x} \int_0^{\infty} (e^s-1-s)^k e^{-\frac{s}{x}}ds ,$$ which clearly produces a rational function: if we expand the term $(e^s-1-s)^k$ and integrate with a linear change of variables we get a finite sum $$\sum_{n\ge0} b(n,k) x^n =\frac{(-1)^k}{x}\sum_{i\ge0, j\ge0\atop i+j\le k} \frac{(-1)^j}{i!j!}\Big(\frac{x}{1-jx}\Big)^{k-i-j+1} ,$$ which is a rational function of the form you suggested in last edit.
Rmk: the above transformation can be applied formally, as all needed identities holds in a formal context; however, for $|x|<1/k$ the convergence of the integrals and series are ensured. $$*$$ [edit] One can show that $\beta_k(x)$ is a rational function, without computing it explicitly. The recursive relation of the coefficients $b(n,k)$, translated into the sequence $\beta_k$ reads: $$(1-kx)\beta_k=x^2(x\beta_{k-1})'$$ for $k\ge1$. Since $\beta_0(x):=1$, it follows by induction that $\beta_k$ are rational functions of the form $$\beta_k(x)=\frac{x^{2k}P_k(x)}{(1-x)^k(1-2x)^{k-1}\cdots(1-kx)},$$ for polynomials $P_k(x)$ of degree $\binom{k}{2}$.