It looks like my first response interprets the problem statement to
use unsigned Stirling numbers of the first kind. We find for signed
ones,
$$S_n = (-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j
(-1)^k 2^{j-k} {2n\choose j} {j\brace k}
{k\brack j+1-n}.$$
With the usual EGFs we get
$$(-1)^{n+1} \sum_{j=n}^{2n} \sum_{k=j+1-n}^j
(-1)^k 2^{j-k} {2n\choose j} j! [z^j] \frac{(\exp(z)-1)^k}{k!}
\\ \times k! [w^k]
\frac{1}{(j+1-n)!} \left(\log\frac{1}{1-w}\right)^{j+1-n}.$$
Now we have
$${2n\choose j} j! \frac{1}{(j+1-n)!}
= \frac{(2n)!}{(2n-j)! \times (j+1-n)!}
= \frac{(2n)!}{(n+1)!} {n+1\choose j+1-n}.$$
This yields for the sum
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!}
\sum_{j=n}^{2n} {n+1\choose j+1-n} 2^j
\\ \times [z^j] \sum_{k=j+1-n}^j (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1-n}
\\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j
\\ \times [z^{n+j}] \sum_{k=j+1}^{j+n} (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1}.$$
Observe that $(\exp(z)-1)^k = z^k + \cdots$ and hence we may extend
the inner sum beyond $j+n$ due to the coefficient extractor
$[z^{n+j}].$ We find
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j
\\ \times [z^{n+j}] \sum_{k\ge j+1} (-1)^k 2^{-k}
(\exp(z)-1)^k [w^k]
\left(\log\frac{1}{1-w}\right)^{j+1}.$$
Furthermore note that $\left(\log\frac{1}{1-w}\right)^{j+1} = w^{j+1}
+\cdots$ so that the coefficient extractor $[w^k]$ covers the entire
series, producing
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^n
\sum_{j=0}^{n} {n+1\choose j+1} 2^j [z^{n+j}]
\left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}.$$
Working with formal power series we are justified in writing
$$[z^{n+j}] \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}
= [z^{n-1}] \frac{1}{z^{j+1}}
\left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}$$
because the logarithmic term starts at $(-1)^{j+1} z^{j+1}/2^{j+1}.$
To see this write
$$-\frac{\exp(z)-1}{2}
+ \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2}
- \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3}
\pm \cdots$$
We continue
$$(-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=0}^{n} {n+1\choose j+1} 2^{j+1}
\frac{1}{z^{j+1}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j+1}
\\ = (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=1}^{n+1} {n+1\choose j} 2^{j}
\frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}.$$
The term for $j=0$ in the sum is one and hence only contributes to
$n=1$ so that we may write
$$-[[n=1]]
+ (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\sum_{j=0}^{n+1} {n+1\choose j} 2^{j}
\frac{1}{z^{j}} \left(\log\frac{1}{1+(\exp(z)-1)/2}\right)^{j}
\\ = -[[n=1]]
+ (-1)^{n+1} \frac{(2n)!}{(n+1)!} 2^{n-1} \\ \times [z^{n-1}]
\left(1+\frac{2}{z}
\log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}.$$
Finally observe that
$$\left(1+\frac{2}{z}
\log\frac{1}{1+(\exp(z)-1)/2}\right)^{n+1}
\\ = \left(1+\frac{2}{z}
\left( -\frac{\exp(z)-1}{2}
+ \frac{1}{2} \frac{(\exp(z)-1)^2}{2^2}
- \frac{1}{3} \frac{(\exp(z)-1)^3}{2^3}
\pm \cdots \right)\right)^{n+1}
\\ = \left( -\frac{1}{4} z - \cdots \right)^{n+1}$$
and furthermore
$$[z^{n-1}] \left((-1)^{n+1}
\frac{1}{4^{n+1}} z^{n+1} + \cdots \right) = 0$$
which is the claim.
We seek to verify that
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
{q\brace m} {n\brack q+k}
= {n-1\choose m} {n-m\brack k}.$$
Using the standard EGFs the LHS becomes
$$\sum_{q=m}^{n-k} (-1)^{q-m} {k-1+q\choose k-1}
q! [z^q] \frac{(\exp(z)-1)^m}{m!}
n! [w^n] \frac{1}{(q+k)!} \left(\log\frac{1}{1-w}\right)^{q+k}
\\ = \frac{n!}{(k-1)! \times m!} [w^n] \sum_{q=m}^{n-k} (-1)^{q-m}
[z^q] (\exp(z)-1)^m
\frac{1}{q+k} \left(\log\frac{1}{1-w}\right)^{q+k}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}]
\sum_{q=m}^{n-k} (-1)^{q-m}
[z^q] (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q+k-1} \frac{1}{1-w}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q=m}^{n-k} (-1)^{q-m}
[z^{q+k-1}] z^{k-1} (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q+k-1}
\\ = \frac{(n-1)!}{(k-1)! \times m!} [w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q=m+k-1}^{n-1} (-1)^{q-(k-1)-m}
[z^{q}] z^{k-1} (\exp(z)-1)^m
\left(\log\frac{1}{1-w}\right)^{q}.$$
Now as $\log\frac{1}{1-w} = w + \cdots$ when $q\gt n-1$ there is
no contribution from the logarithmic power term due to the coefficient
extractor $[w^{n-1}]$ so we find
$$(-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times \sum_{q\ge m+k-1} (-1)^{q}
\left(\log\frac{1}{1-w}\right)^{q}
[z^{q}] z^{k-1} (\exp(z)-1)^m.$$
Note that $z^{k-1} (\exp(z)-1)^m = z^{m+k-1} + \cdots$ which
means that the remaining sum / coefficient etractor pair covers the
entire series and we get
$$(-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times (-1)^{k-1} \left(\log\frac{1}{1-w}\right)^{k-1}
\left(\exp\left(-\log\frac{1}{1-w}\right)-1\right)^m
\\ = (-1)^{m+(k-1)} \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1}] \frac{1}{1-w}
\\ \times (-1)^{k-1} \left(\log\frac{1}{1-w}\right)^{k-1} (-w)^m
\\ = \frac{(n-1)!}{(k-1)! \times m!}
[w^{n-1-m}] \frac{1}{1-w}
\left(\log\frac{1}{1-w}\right)^{k-1}
\\ = \frac{(n-1)!}{m!}
[w^{n-1-m}] \frac{1}{1-w}
\frac{1}{(k-1)!} \left(\log\frac{1}{1-w}\right)^{k-1}
\\ = \frac{(n-1)!}{m!}
(n-m) [w^{n-m}]
\frac{1}{k!} \left(\log\frac{1}{1-w}\right)^{k}
\\ = \frac{(n-1)!}{m! \times (n-1-m)!}
(n-m)! [w^{n-m}]
\frac{1}{k!} \left(\log\frac{1}{1-w}\right)^{k}
\\ = {n-1\choose m} {n-m\brack k}.$$
This is the claim.
Best Answer
We seek to show that with $0\le k\le n$ the following identity holds:
$$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} = \sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$
We will start with the LHS. The chapter 6.2 on Eulerian Numbers of Concrete Mathematics by Knuth et al. proposes the formula
$${n\brace m} = (-1)^{n-m+1} \frac{n!}{(m-1)!} \sigma_{n-m}(-m)$$
where $\sigma_n(x)$ is a Stirling polynomial and we have the identity
$$\left(\frac{1}{z} \log\frac{1}{1-z}\right)^x = x \sum_{n\ge 0} \sigma_n(x+n) z^n.$$
We get
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^x = x \sigma_{n-m}(x+n-m)$$
and hence
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n} = -n \sigma_{n-m}(-m)$$
which implies that for $n\ge m\ge 1$
$$\bbox[5px,border:2px solid #00A000]{ {n\brace m} = (-1)^{n-m} \frac{(n-1)!}{(m-1)!} [z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}.}$$
This gives for the LHS
$$\sum_{j=1}^k (-1)^{k-j} {2n+1\choose k-j} (-1)^n \frac{(n+j-1)!}{(j-1)!} [z^n] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-j} \\ = (-1)^{n-k+1} n! [z^n] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1} [w^{k-1}] (1+w)^{2n+1} \\ \times \sum_{j=1}^k {n+j-1\choose n} (-1)^{j-1} w^{j-1} \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-j+1}.$$
Now the coefficient extractor in $w$ enforces the upper limit of the sum and we may extend $j$ to infinity, getting
$$(-1)^{n-k+1} n! [z^n] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1} [w^{k-1}] (1+w)^{2n+1} \frac{1}{(1+w/(\frac{1}{z} \log\frac{1}{1-z}))^{n+1}} \\ = (-1)^{n-k+1} n! [z^n] [w^{k-1}] (1+w)^{2n+1} \frac{1}{(w+\frac{1}{z} \log\frac{1}{1-z})^{n+1}}.$$
Continuing,
$$ (-1)^{n-k+1} n! [z^n] [w^{n+k}] (1+w)^{2n+1} \frac{1}{(1+\frac{1}{w} \frac{1}{z} \log\frac{1}{1-z})^{n+1}} \\ = (-1)^{n-k+1} n! [z^n] [w^{n+k}] (1+w)^{2n+1} \sum_{q\ge 0} {n+q\choose n} (-1)^q \frac{1}{w^q} \left(\frac{1}{z} \log\frac{1}{1-z}\right)^q \\ = (-1)^{n-k+1} n! [z^n] \sum_{j=n+k}^{2n+1} {2n+1\choose j} {n+j-(n+k)\choose n} (-1)^{j-(n+k)} \\ \times \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j-(n+k)} \\ = (-1)^{n-k+1} n! [z^n] \sum_{j=0}^{n-k+1} {2n+1\choose j+n+k} {n+j\choose n} (-1)^{j} \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j} \\ = (-1)^{n-k+1} n! \sum_{j=0}^{n-k+1} {2n+1\choose j+n+k} {n+j\choose n} (-1)^{j} [z^{n+j}] \left(\log\frac{1}{1-z}\right)^{j} \\ = (-1)^{n-k+1} n! \sum_{j=0}^{n-k+1} {2n+1\choose j+n+k} {n+j\choose n} (-1)^{j} \\ \times \frac{j!}{(n+j)!} \times (n+j)! [z^{n+j}] \frac{1}{j!} \left(\log\frac{1}{1-z}\right)^{j} \\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1} {2n+1\choose j+n+k} (-1)^j {n+j\brack j} \\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1} {2n+1\choose 2n-j+1} (-1)^{n-k-j+1} {2n-k-j+1\brack n-k-j+1} \\ = \sum_{j=0}^{n-k+1} {2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
The Stirling number is zero for $j=n-k+1$ and we get at last
$$\sum_{j=0}^{n-k} {2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
This is the RHS and we have the claim.