Hint: We could look at the situation with the help of exponential generating functions.
The following holds: If $$A(z)=\sum_{k=0}^\infty a_k \frac{z^k}{k!}$$ is an exponential generating function of the sequence $(a_k)$, then
\begin{align*}
A(z)e^z=\sum_{k=0}^\infty \left(\sum_{j=0}^k\binom{k}{j} a_j \right)\frac{z^k}{k!}
\end{align*}
and applying the differential operator $D_z=\frac{d}{dz}$ we obtain
\begin{align*}
D_z A(z)=\sum_{k=0}^\infty a_{k+1}\frac{z^k}{k!}
\end{align*}
Since the exponential generating function of the Bell numbers is
\begin{align*}
B(z)=\exp\left({e^z-1}\right)=\sum_{n=0}^\infty B_n\frac{z^n}{n!}
\end{align*}
the relation
\begin{align*}
\ \qquad\qquad\sum_{j=0}^k\binom{k}{j}B_j=B_{k+1}\qquad\qquad k\geq 0
\end{align*}
translates into
\begin{align*}
B(z)e^z&=D_z B(z)\\
\end{align*}
We could now try to find an exponential generating function $P(z)$ for the partial ordinary Bell polynomials. The expression
\begin{align*}
\sum_{j=0}^k(-1)^{j-1}\binom{k}{j}\hat{B}_{m,j}(x_1,x_2,\ldots,m-j+1)
\end{align*}
looks like the coefficient of
\begin{align*}
(-1)^{k-1}P(z)e^{-z}
\end{align*}
We could now check if this series could also be represented using operators like $D_z, e^z$, etc. Here we use the notation $\hat{B}_{m,j}$ in accordance with the Wiki page.
[Add-on 2017-04-30] Some information regarding ordinary generating functions of partial ordinary Bell polynomials. We use the coefficient of operator $[t^n]$ to denote the coefficient of $t^n$ of a series $A(t)$.
According to the referred Wiki page are the partial ordinary Bell polynomials
\begin{align*}
\hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1})
\end{align*}
the coefficients of a bivariate generating function $\hat{\Phi}(t,u)$ with
\begin{align*}
\hat{\Phi}(t,u)&=\exp\left(u\sum_{l=1}^\infty x_lt^l\right)\\
&=\sum_{k=0}^\infty\left(\sum_{l=1}^\infty x_lt^l\right)^k\frac{u^k}{k!}\\
&=\sum_{k=0}^\infty\left(\sum_{n=k}^\infty\hat{B}_{n,k}\left(x_1,x_2,\ldots,x_{n-k+1}\right)t^n\right)\frac{u^k}{k!}
\end{align*}
We can write
\begin{align*}
\hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1})&=[t^mu^j]\hat{\Phi}(t,u)\\
&=[t^m]\underbrace{\left(\sum_{l=1}^\infty x_lt^l\right)^j}_{:= \hat{B}_j(t)}\tag{1}
\end{align*}
We recall the Euler transformation formula of a series
\begin{align*}
A(t)=\sum_{n= 0}^\infty a_nt^n\qquad\qquad
\frac{1}{1-t}A\left(\frac{-t}{1-t}\right)=\sum_{n= 0}^\infty \left(\sum_{j=0}^m\binom{m}{j}(-1)^ja_j\right)t^n\tag{2}
\end{align*}
This transformation formula together with a proof can be found e.g. in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev. See remark 2 with $\lambda=1, \mu=-1$.
We conclude according to (1) and (2)
\begin{align*}
\sum_{j=0}^m(-1)^{j}\binom{m}{j}\hat{B}_{m,j}(x_1,x_2,\ldots,x_{m-j+1})
&=[t^mu^j]\frac{1}{1-t}\hat{\Phi}\left(\frac{-t}{1-t},u\right)\\
&=[t^m]\frac{1}{1-t}\hat{B}_j\left(\frac{-t}{1-t}\right)\tag{3}
\end{align*}
Note: Regrettably there seems to be no simple representation of the coefficients in (3) based on the ordinary generating function. An exponential generating function
\begin{align*}
\sum_{n=k}^\infty\hat{B}_{n,k}\left(x_1,x_2,\ldots,x_{n-k+1}\right)\frac{t^n}{n!}
\end{align*}
of the partial ordinary Bell polynomials is not stated at the Wiki-page about Bell polynomials and useful represenations are not known to me.
