Are products of analytic continuations also analytic?

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The question of the value, if any depending on which answer you choose, of $\sum_{n=1}^\infty n$ has been addressed a few times. At least here Does $\zeta(-1)=-1/12$ or $\zeta(-1) \to -1/12$? and here Why does $1+2+3+\cdots = -\frac{1}{12}$?. I do not want to re-open that question in general, but I have a question about a specific step of one of the approaches (or purported approaches as you may like) to computing the result.

Under the zeta function regularization technique, one ultimately observes that $$ \left( 1 - 2^{1-s} \right) \zeta(s) = \eta(s) $$ for the Riemann zeta function $\zeta$ and the Dirichlet eta function $\eta$. One usually arrives at this result by using the series representations of these two functions and performing manipulations on them that are valid for complex values of $s$ where the series representations of $\zeta$ and $\eta$ converge.

That seems fine as far as it goes, under the assumption that each function is evaluated at a value of $s$ where the series converges. The method then continues to assert that the relationship holds for the analytic continuations of $\zeta$ and $\eta$. That's the step that motivates my question.

Is it generally true that if $f(s) g(s) = h(s)$ on an open set $U$ that this relationship will continue to hold for their analytic continuations to larger sets? If not generally true, what is the special property of $\zeta$ and $\eta$ that makes it true for the case outlined above?

My sense is that it's not generally true because of differences in which potential supersets of $U$ each individual function has an analytic continuation, but I'm operating well on the fringe of my understanding of this topic.

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Yes, it is true, by the identity theorem, as stated by e.g., Wikipedia:

Given functions $f$ and $g$ holomorphic on a domain $D$ (open and connected subset), if $f = g$ on some $S\subseteq D$, $S$ having an accumulation point, then $f = g$ on $D$.

In particular, $f(s), g(s), h(s)$ are analytic functions and $f(s)g(s) = h(s)$ on any open set $U\subseteq D$ (or indeed, any set $S$ that has a limit point), then $f(s)g(s) = h(s)$ on the whole set $D$.

You may gain some intuition on the identity theorem by expecting that analytic functions behave, to some extent, like high-degree polynomials - which may be expected since they have power series representations. Any two degree-$n$ polynomials are identical if they agree on any $n+1$ points. Similarly, any two analytic functions are identical if they agree on any infinite set of points - with the important caveats that the set has a limit point, and that the domain they are defined on is connected.