Let $A(x)$ be the exponential generating function for $a_n$, i.e. $\frac{d^n}{dx^n}A(0)=a_n$, and let $B(x)=A(\log(\frac{1}{1-x}))$.
We need to show that $\frac{d^n}{dx^n}B(0)=b_n$.
Let us find an expression for $\frac{d^n}{dx^n}B(x)$ using Stirling coefficients: $\frac{d^n}{dx^n}B(x)=\sum_{k=1}^n\left[\matrix{n\\k}\right]a_k^n(x)A^{(k)}(-\log(1-x)).$
We have \frac{d^{n+1}}{dx^{n+1}}B(x)=\sum_{k=1}^n\left[\matrix{n \\ k}\right]\left(a_k^{n'}(x)A^{(k)}(-\log(1-x))+\frac{a_k^n(x)}{1-x}A^{(k+1)}(-\log(1-x))\right)=a_1^{n'}(x)A^{(1)}(-\log(1-x))+\frac{a_n^n(x)}{1-x}A^{(n+1)}(-\log(1-x))+\sum_{k=2}^n\left(\left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}\right)A^{(k)}(-\log(1-x))=\sum_{k=1}^{n+1}\left(\left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}\right)A^{(k)}(-\log(1-x))
So we need $a_k^n(x)$ to be such that \left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}=\left[\matrix{n+1 \\ k}\right]a_k^{n+1}(x). Also, we need $a_1^1(x)=\frac{1}{1-x}$.
Note that $n\left[\matrix{n \\ k}\right]+\left[\matrix{n \\ k-1}\right]=\left[\matrix{n+1 \\ k}\right]$. Therefore, we obtain the following expressions: $a_k^n(x)=(1-x)^{-n}$ and $\frac{d^n}{dx^n}B(x)=\frac{1}{(1-x)^n}\sum_{k=1}^n\left[\matrix{n\\k}\right]A^{(k)}(-\log(1-x)).$
Now just plug in $x=0$ into the final expression.