An operator performing the mapping is $$O= D/(e^D-1)=e^{B.(0)D},$$
where $D=d/dx $ and $(B.(0))^n=B_n(x)|_{x=0}$, since the Bernoulli polynomials are an Appell sequence.
Edit (6/20/2017):
This operator is essentially the Todd operator. See the discussions on pg. 30 and Appendix B of "Permutohedra, associahedra, and beyond" by Postnikov of the Todd operator as a transform of the homogeneous volume polynomials for classes of polytopes into a generalized Ehrhart polynomial coding the number of lattice points in the polytopes.
(Edit 8/2018)
For some idea of the importance of this Todd operator in modern mathematics and physics, see New Models for Veneziano Amplitudes: Combinatorial, Symplectic and Supersymmetric Aspects by Kholodenko.
(Edit 9/2024)
A rep of the Bernoulli op that is more robust than
$$T_{x \to (x+b.)}= e^{b.\partial_x}=: \frac{\partial_x}{e^{\partial_x}-1}$$
is
$$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$
Define the Bernoulli translation op $T_{x \to (x+b.)}$ as the op that umbrally translates the variable $x$ to $(x+b.) = B.(x)$ such that
$$T_{x \to (x+b.)} x^n = (x+b.)^n = \sum_{k=0}^n \binom{n}{k}(b.)^k x^{n-k}= \sum_{k=0}^n \binom{n}{k}b_k x^{n-k} = B_n(x)$$
where $b_n$ is the $n$-th Bernoulli number and $B_n(x)$ the $n$-th Bernoulli polynomial.
Then a diff op rep when acting on $x^n$ is
$$T_{x \to (x+b.)} = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1}.$$
A more robust, extended rep of the Bernoulli op that allows action on $x^s$ where $s$ is complex can be realized via an operator compositional-inverse pair.
With $D = \frac{\partial}{\partial x} = \partial_x$, Define the finite diff op $\delta_x$ by
$$\delta_x x^s = (x+1)^s-x^s = (T_{x \to x+1}-1) x^s.$$
Then its diff op rep is (by analytic continuation)
$$\delta_x x^s = (e^{\partial_x}-1) x^s = (x+1)^s -x^s$$
and inverting
$$\partial_x = \ln(1+\delta_x),$$
so when acting on $x^n$
$$T_{x \to (x+b.)=B.(x)} = e^{b.\partial_x} = \frac{\partial_x}{e^{\partial_x}-1} = \frac{\partial_x}{\delta_x} = \frac{\ln(1+\delta_x)}{\delta_x}$$
$$ = \sum_{k \geq 0} (-1)^k \frac{1}{k+1} \delta_x^k = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} e^{k\partial_x}$$
$$ = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$
Removing the intermediate steps gives the Bernoulli translation rep
$$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}$$
with action
$$T_{x \to (x+b.)}x^s = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = -s\zeta(-s+1,x),$$
essentially the Helmut Hasse formula for the Hurwitz zeta function $\zeta(s,x)$. Taking this as the interpretation of
$$T_{x \to (x+b.)}x^s = (x+b.)^s = (B.(x))^s = B_s(x)$$
gives the Bernoulli function
$$B_s(x) = -s\zeta(-s+1,x) = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = T_{x \to (x+b.)}x^s,$$
and as
$$T_{x \to (x+b.)} \frac{(x+1)^{n+1}-x^{n+1}}{n+1} = e^{b.\partial_x} \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$$
$$ = \frac{(b.+x+1)^{n+1}-(b.+x)^{n+1}}{n+1}$$
$$ = \frac{(B.(x+1))^{n+1}-(B.(x))^{n+1}}{n+1} = \frac{B_{n+1}(x+1)-B_{n+1}(x)}{n+1} $$
$$= \frac{\partial_x}{\delta_x}\delta_x \frac{ x^{n+1}}{n+1} = \partial_x \frac{ x^{n+1}}{n+1} = x^n$$
so does
$$T_{x \to (x+b.)} \frac{(x+1)^{s+1}-x^{s+1}}{s+1} = \frac{B_{s+1}(x+1)-B_{s+1}(x)}{s+1}$$
$$ = - \zeta(-s,x+1) -(-\zeta(-s,x)) = x^s = \partial_x \frac{x^{s+1}}{s+1}.$$
Note the action of the inverse op
$$ \frac{\delta_x}{\partial_x}x^n = \frac{e^{\partial_x}-1}{\partial_x}x^n = \sum_{k \geq 0} \frac{1}{k+1} \frac{\partial_x^k}{k!}x^n = \sum_{k \geq 0} \binom{n}{k} \frac{1}{k+1} x^{n-k}= \int_0^1 (x+t)^ndt $$
$$= \int_{x}^{x+1}t^n dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1} $$
has a rep as a sliding average, which, with $\hat{b}_k = \frac{1}{k+1}$, gives the action
$$T_{ x \to (x+\hat{b}.) = \hat{B}.(x)}x^s = \int_{x}^{x+1}t^s dt = \int_0^1 (x+t)^sdt = \frac{(x+1)^{s+1}-x^{s+1}}{s+1},$$
with a log function in the limiting case $s \to -1$.