sequences and series - Are products of analytic continuations also analytic?

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.

Answer

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.