Analytic continuation of $Phi(s)=sum_{n ge 1} e^{-n^s}$

This is currently a partial answer, refining the idea given by @reuns.

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The series $Phi(s)=sum_{n=1}^infty e^{-n^s}$ converges iff $s>0$ is real. Using the Cahen–Mellin integral $$e^{-x}=frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}Gamma(z)x^{-z},dzqquad(x,c>0)$$ with $x=n^s$ and $c>1/s$, we get $$Phi(s)=frac{1}{2pi i}int_{c-iinfty}^{c+iinfty}Gamma(z)zeta(sz),dz.$$

For $0<s<1$, the integrand tends to $0$ rapidly enough when $ztoinfty$ in the half-plane $Re zleqslant c$ and out of a neighborhood of the line $L={z : Im z=0wedgeRe zleqslant 1/s}$. This allows us to deform the path of integration, making it encircle $L$, and we see that $Phi(s)$ is equal to the (infinite) sum of residues of the integrand at its poles (which are $z=1/s$ and $z=-n$ for nonnegative integers $n$). Computing these, we get $$Phi(s)=Gammaleft(1+frac1sright)+sum_{n=0}^inftyfrac{(-1)^n}{n!}zeta(-ns).tag{*}label{theseries}$$

This series converges for complex $sneq 0$ with $Re s<1$ (at least; the singularities at $s=-1/n$ for $ninmathbb{Z}_{>0}$ are removable), and gives the analytic continuation of $Phi(s)$ in this region.

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Update. In fact the series eqref{theseries} converges also for some $s$ with $Re s=1$. This (as well as the convergence for $Re s<1$) is seen using the functional equation for $zeta$: $$zeta(-ns)=-(2pi)^{-ns}frac1pisinfrac{npi s}{2}Gamma(1+ns)zeta(1+ns)$$ and Stirling's asymptotics for $Gamma$. At $s=1+it$, as $ntoinfty$, the latter gives $$frac1{n!}Big|Gamma(1+ns)Big|asymp(1+t^2)^{(2n+1)/4}e^{-ntarctan t},$$ so that $$frac1{n!}Big|zeta(-ns)Big|asymp(2pi)^{-n-1}(1+t^2)^{(2n+1)/4}e^{ntarctan 1/t}.$$

Thus, eqref{theseries} converges at $s=1+it$ with $|t|<tau$, where $tauapprox 2.24$ satisfies $$(1+tau^2)e^{2tauarctan 1/tau}=4pi^2.$$

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The remaining question is whether we can extend this region further.