Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive relationship of Bernoulli numbers and induction, but unsuccessful. Now I'm wondering if this is a viable proof strategy. Any comments are appreciated!
Edit: This is a conjecture, I evaluated it numerically for $n,m\leq 20$.

With the work by @RenéGy and @epi163sqrt we are trying to prove the
statement
$$(-1)^n \sum_{g=0}^m \frac{B_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{B_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{1}{n+m+1} {n+m\choose m}^{-1}$$
We prove this for $n\ge m$, it then follows by symmetry for $m\ge n.$
Using
$$B_n = (-1)^n \sum_{k=0}^n {n\choose k} B_k$$
we get for the first piece
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n+g+1} {n+g+1\choose k} B_k.$$
Extracting $k=0$ we get
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g} (-1)^{g+1}
= \sum_{g=0}^m \frac{1}{n+m-g+1} {m\choose g} (-1)^{m-g+1}
\\ =  [z^{n+m+1}] \log\frac{1}{1-z}
\sum_{g=0}^m z^g {m\choose g} (-1)^{m-g+1}
\\ = (-1)^{m+1} [z^{n+m+1}] \log\frac{1}{1-z}
(1-z)^m
= - [z^{n+m+1}] \log\frac{1}{1-z} (z-1)^m.$$
Recall from MSE
4316307
that with $1\le k\le n$
$${n\choose k}^{-1}
= k [z^n] \log\frac{1}{1-z}
(z-1)^{n-k}.$$
We put $n := n+m+1$ and $k := n+1$ to get
$$-\frac{1}{n+1} {n+m+1\choose n+1}^{-1}
= -\frac{1}{n+m+1} {n+m\choose n}^{-1}.$$
This also could have been obtained by summing residues of a suitable
function. Good, we have the RHS. Now we get for the remainder
$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g}
(-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g+1\choose k} B_k
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g\choose k-1} \frac{B_k}{k}
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1}.$$
We get two components, the first is,
$$\sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{m-1} {n+g\choose k} \frac{B_{k+1}}{k+1}
\\ = \sum_{k=0}^{m-1} \frac{B_{k+1}}{k+1}
\sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose k}.$$
The inner sum is
$$[z^k] (1+z)^n \sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g
= - [z^k] (1+z)^n (-1)^m z^m = 0$$
since $k\lt m$. That leaves
$$\sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=m}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1}
\\ = \sum_{g=0}^m {m\choose g}
(-1)^{g+1} \sum_{k=0}^{n-m+g} {n+g\choose m+k} 
\frac{B_{m+k+1}}{m+k+1}
\\ = \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1}
\sum_{g=k+m-n}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$
Now when $n+g\lt m+k$ or  $g\lt m-n+k$ the second binomial coefficient
is zero, so we may lower $g$  to zero (observe that the first binomial
coefficient is zero when $g\lt 0$ so we also may raise to zero when $k+m-n \lt 0$):
$$\sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1}
\sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$
The inner sum is
$$[z^{m+k}] (1+z)^n
\sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g
= - [z^{m+k}] (1+z)^n (-1)^m z^m
\\ = - (-1)^m [z^k] (1+z)^n = - (-1)^m {n\choose k}.$$
We have obtained
$$- (-1)^m \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1} {n\choose k},$$
which is minus the second piece and concludes the argument.
Addendum. Obviously what we have proved here is with $b_n=
(-1)^n \sum_{k=0}^n {n\choose k} a_k$ then
$$(-1)^n \sum_{g=0}^m \frac{b_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$
We get for an ordinary binomial transform $b_n = \sum_{k=0}^n {n\choose
k} a_k$ the relation
$$\sum_{g=0}^m (-1)^{g+1} \frac{b_{n+g+1}}{n+g+1} {m\choose g}
+ (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g}
= - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$
