Equations of this type are known. You may see, for example, the classical
book "Primzahlen" by Landau paragraph 67. "Continuation of the zeta function by partial integration"
There it is proved the formula
$$ (s-1)(\zeta(s)-1)-1=-\frac{(s-1)s}{2!}(\zeta(s+1)-1)-
\frac{(s-1)s(s+1)}{3!}(\zeta(s+2)-1)-\cdots$$
$$\cdots-\frac{(s-1)s\cdots(s+q)}{(q+2)!}(\zeta(s+q+1)-1)
-\frac{(s-1)s\cdots(s+q+1)}{(q+2)!}\int_1^\infty \frac{\rho(x)^{q+2}}{x^{s+q+2}} dx.$$
From this we may get the beautiful formula
$$1=\binom{s-1}{1}(\zeta(s)-1)+\binom{s}{2}(\zeta(s+1)-1)+\binom{s+1}{3}(\zeta(s+2)-1)+
\cdots$$
(See Titchmarsh 2.14, formula (2.14.1) and also (2.14.2))
From the formula in Landau you see that the integral
$\int_1^\infty \frac{\rho(x)^r}{x^{s+r}}\,dx$ can be expressed in terms of the
$\zeta(s+k)$.
Here is a cheaper alternative depending on what you mean by a modification. Consider $L(z)$ any Dirichlet L function different from $\zeta$.
Joint universality theorem:
Let $K$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$ with connected complement. For any two functions $f_1$ and $f_2$ holomorphic in the interior of $K$ (vanishing or not) and every $\epsilon>0$, we have that the limit
$$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \{ t \leq T: \sup |f_1(z) - \log \zeta(z +i t)| + \sup |f_2(z) - \log L(z +i t)| < \epsilon\} $$
is positive for $\lambda$ being the Lebesgue measure.
From this, we can deduce:
Corollary: Let $K_0$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$.
Let $f$ be a continuous function on $K_0$, which is holomorphic on an open set containing $K_0$. For every $\epsilon_0>0$, we have that the limit
$$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big\{ t \leq T: \sup\limits_{z \in K_0} \left| f(z) - \frac{\log \zeta(z +i t)}{\log L(z+ it)}\right| < \epsilon_0\Big\} $$
is positive for $\lambda$ being the Lebesgue measure.
Proof:
By Runge's theorem, it is sufficient to approximate rational functions, whose poles lie outside of $K_0$. Let $p(z)$ and $q(z)$ be polynomials such that $q$ does not vanish on $K_0$. Consider $\epsilon_0>0$ sufficiently small (to be made precise as we go on).
Let $K :=\mathbb{C}-O$, where $O$ is the unbounded, connected component of $\mathbb{C}-K_0$. Consider $\epsilon>0$ sufficiently small, then use the joint universality theorem for $f_1(z)=p(z)$ and $f_2(z) =q(z)$.
We want to show that
$$\sup | f_1/f_2(z) - \frac{\log \zeta}{\log L}(z+i t) |< \epsilon_0.$$
We estimate the left-hand side:
$$ \leq \sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | + \sup | \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)|.$$
The first summand is easy to estimate:
$$\sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | \leq \sup_{z \in K_0} \left| f_2(z)^{-1} \right| \epsilon.$$
The second one is a little bit harder:
$$ \sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big| \leq $$
$$ \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big| \sup | \log L(z+i t) -f_2(z) | < \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z) \log L(z+i t)} \Big| \epsilon,$$
because we have to estimate
$$ \sup | \frac{\log \zeta}{\log L}(z+i t) | $$
uniformly in $t$.
This is indeed possible, we have that
$$\sup | f_2(z) | - \sup | \log L(z + i t) | < \epsilon$$
and
$$ \sup | \log \zeta(z + i t) | - \sup | f_1(z) | < \epsilon$$
by the reversed triangle inequality.
So for $\epsilon \leq \sup | f_2(z) |/2$ and $\epsilon \leq \sup | f_1(z) |$ , we have that
$$ \sup | \log L(z + i t) | > \sup | f_2(z) |/2$$
and
$$ \sup | \log \zeta(z + i t) | < 2 \sup | f_1(z) | .$$
So
$$\epsilon_0 := \max\{ \frac{1}{2} \sup |f_2^{-1}| \epsilon, \frac{1}{2} 4* \sup |f_1f_2^{-2}| \epsilon \}$$
will do.
This finishes the proof of the corollary assuming the Joint universality theorem.
Best Answer
When I was writing my answer, Dan Romik answer appeared. Mine is the same but with more detail.
We have $$\log\Gamma(1-x)=\gamma x+\sum_{n=2}^\infty\zeta(n)\frac{x^n}{n}$$ (this is known and is also an exercise in complex analysis). Hence by differentiation $$-\frac{\Gamma'(1-x)}{\Gamma(1-x)}-\gamma=\sum_{n=2}^\infty \zeta(n) x^{n-1}$$ It follows that (I change $\kappa$ into $x$) $$\sum_{n=1}^\infty\frac{x^n-1}{(x+1)^n}\zeta(n+1)= \frac{\Gamma'(1-\frac{1}{x+1})}{\Gamma(1-\frac{1}{x+1})}+\gamma -\frac{\Gamma'(1-\frac{x}{x+1})}{\Gamma(1-\frac{x}{x+1})}-\gamma$$ $$=\frac{\Gamma'(\frac{x}{x+1})}{\Gamma(\frac{x}{x+1})}-\frac{\Gamma'(\frac{1}{x+1})}{\Gamma(\frac{1}{x+1})}=\frac{\Gamma'(y)}{\Gamma(y)}-\frac{\Gamma'(1-y)}{\Gamma(1-y)}=\frac{d}{dy}\log(\Gamma(y)\Gamma(1-y))=\frac{d}{dy}\log\frac{\pi}{\sin\pi y}$$ $$=-\frac{\pi\cos\pi y}{\sin\pi y}=-\pi\cot\pi y=\pi\cot\frac{\pi x}{x+1}$$ $$=-\pi\tan\Bigl(\frac{\pi}{2}-\frac{\pi x}{x+1}\Bigr)=-\pi\tan\Bigl(\frac{\pi}{2}\frac{1-x}{1+x}\Bigr)=\pi\tan\Bigl(\frac{\pi}{2}\frac{x-1}{x+1}\Bigr).$$