$\underline{\smash{\textbf{Background:}}}$ It is a well-known fact that the original definition of the zeta function, namely $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} $$ is only defined when $\operatorname{Re}(s)>1$. We can then extend the domain by redefining $$ \zeta(s) = \frac{\eta(s)}{1-2^{1-s}} $$ whenever $2^{1-s} \neq 1$. Here, $\eta(s)$ is the alternating zeta function, which converges for all $s$ such that $\operatorname{Re}(s)>0$.
Now we extend the domain even further to the entire complex plane without $\operatorname{Re}(s) = 1$ by using the symmetry identity $$ \zeta(s) = 2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s), $$ which holds for all $s$ in the strip $0< \operatorname{Re}(s)<1$ and defining the zeta function using this identity for $\operatorname{Re}(s) < 0$.
$\underline{\smash{\textbf{My question:}}}$ How can this analytical continuation be defined for $s$ with real part zero? Applying the last identity to such numbers define $\zeta(s)$ in terms of $\zeta(1-s)$, but $1-s$ has a real part equal to $1$. But we excluded these numbers in the definition (and first continuation) of the zeta function.