= Solution
Each finite-interval integral $f_n(s)$ is entire in $s$: the integrand is $ (x-n)e^{-(s+1)\log x}$ and its complex derivatives may be integrated on the compact interval. On a compact subset of $\operatorname{Re}s>0$, choose $\delta>0$ with $\operatorname{Re}s\ge\delta$. Then $|f_n(s)|\le n^{-1-\delta}$. The Weierstrass uniform-convergence theorem gives convergence and holomorphy of $\sum_nf_n(s)$ throughout that half-plane.
<Integration by parts> gives
$$
sf_n(s)=-(n+1)^{-s}+\int_n^{n+1}x^{-s}\,dx.
$$
For $\operatorname{Re}s>1$, sum absolutely and use the integral over $[1,\infty)$ to obtain $s\sum_nf_n(s)=1-\zeta(s)+1/(s-1)$. Rearranging gives the first requested identity.
On $[n,n+1]$, the magnitude of $(x-n)(x-n-1)$ is at most $1/4$. On a compact subset of $\operatorname{Re}s>-1$, its integrals are bounded by a convergent constant multiple of $n^{-1-\delta}$ for some $\delta>0$. Thus $\sum_ng_n(s)$ is <holomorphic> there. The prefactor $s(s+1)$ is entire, leaving only the explicit <simple pole> at one in $\widetilde\zeta$, with <residue> one.
Differentiate $(x-n)(x-n-1)x^{-s-1}$ and integrate. Its endpoint values both vanish, giving
$$
\boxed{(s+1)g_n(s)-2f_n(s)=-\int_n^{n+1}x^{-s-1}\,dx=\frac{(n+1)^{-s}-n^{-s}}s.}
$$
At $s=0$ the last quotient has its continuous value $-\log((n+1)/n)$. For $\operatorname{Re}s>1$, telescoping over $n$ gives $(s+1)\sum_ng_n-2\sum_nf_n=-1/s$. Substitution into $\widetilde\zeta$ yields
$$
\widetilde\zeta(s)=\frac1{s-1}+1-s\sum_nf_n(s)=\frac s{s-1}-s\sum_nf_n(s)=\zeta(s).
$$
This supplies the <analytic continuation> to $\operatorname{Re}s>-1$ apart from the <pole>. Evaluating its <holomorphic> expression at zero makes the series term vanish and gives $\boxed{\zeta(0)=-1/2}$.
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