The first integral can be simplified using the property of the Bernoulli polynomials
\begin{equation}
B_{n}\left(0\right)=(-1)^{n}B_{n}\left(1\right)=B_{n}
\end{equation}
where $B_n$ are the Bernoulli numbers. Moreover, for $n=1,2,\cdots$
\begin{equation}
B_{2n+1}=0
\end{equation}
and $B_1=-1/2$. Then, for $n=2p+1$, an odd integer,
\begin{align}
I_{2p+1}&=2\int_{0}^{\infty}\left(\sum_{k=0}^{2p+1}\frac{\left(-1\right)^{k}B_{k}(1)}{k!}x^{k-2p-2}-\frac{1}{x^{2p+1}\left(e^{x}-1\right)}\right)\,dx\\
&=2\int_{0}^{\infty}\left(\frac1x\sum_{k=0}^{2p+1}\frac{B_{k}}{k!}x^{k}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}\\
&=2\int_{0}^{\infty}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}
\end{align}
By changing $x=-y$ in the integral, we have also
\begin{align}
I_{2p+1}&=-2\int_{-\infty}^0\left(-\frac12-\frac1y\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}y^{2r}-\frac{1}{e^{-y}-1}\right)\,\frac{dy}{y^{2p+1}}\\
&=-2\int_{-\infty}^0\left(\frac12-\frac1y\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}y^{2r}+\frac{1}{e^{y}-1}\right)\,\frac{dy}{y^{2p+1}}
\end{align}
as
\begin{equation}
\frac{1}{e^{-y}-1}=-1-\frac{1}{e^{y}-1}
\end{equation}
Thus, by taking the half-sum of both representations
\begin{equation}
I_{2p+1}=\int_{-\infty}^{\infty}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x^{2p+1}}
\end{equation}
To evaluate this integral with the residue method, we close the contour by the upper half circle. At $x=0$, the integrand
\begin{equation}
f(x)=x^{-2p-1}\left(-\frac12+\frac1x\sum_{r=0}^{p}\frac{B_{2r}}{(2r)!}x^{2r}-\frac{1}{e^{x}-1}\right)
\end{equation}
has a removable singularity. Indeed, from the EGF of the Bernoulli numbers, for $x\to0$,
\begin{equation}
f(x)=O\left( x \right)
\end{equation}
The contribution of the half-circle vanishes as for $x\to\infty$
\begin{equation}
f(x)=O\left( x^{-2} \right)
\end{equation}
The poles in the contour are at $x=2ik\pi$, with $k=1,2,\cdots$. Their residues are $-\frac{1}{(2ik\pi)^{2p+1}}$. Finally,
\begin{align}
I_{2p+1}&=2i\pi\sum_{k=1}^\infty\frac{-1}{(2ik\pi)^{2p+1}}\\
&=\frac{1}{(2\pi)^{2p}}\zeta(2p+1)
\end{align}
as expected.
The same method applies for the second integral by remarking that (DLMF)
\begin{equation}
E_{n}\left(1\right)=\frac{2}{n+1}(2^{n+1}-1)B_{n+1}
\end{equation}
one can extend the integral over $(-\infty,\infty)$. Here again, the singularity at the origin is removable due to the EGF of the Euler polynomials. The poles are at $x=(2k+1)i\pi$ and considering that
\begin{equation}
\sum_{k=0}^\infty\frac{1}{(2k+1)^{2p+1}}=\left( 1-2^{-2p-1} \right)\zeta(2p+1)
\end{equation}
the identity follows.
