There are some thoughts just come across my mind about proving $\zeta(s)$ has no poles or essential singularity on $\Re (s) = 1$. So due to the fact that $\zeta(s)$ has an analytic continuation over $\Re (s) > 0$ through the relation with Dirichlet Eta function:
$$\zeta (s) = \frac{\eta (s)}{1- 2^{1-s}}, \Re (s) > 0 \tag{1}$$
Assume that $\zeta(s)$ has another isolated-singularity, denoted as $z_0$ where $\Re(z_0) = 1$, that is different from the zeros. Therefore, $(1)$ implies that $z_0$ is also a singularity of $\eta(s)$.
Since the singularity is isolated, there exists a neighborhood $V$ of $z_0$ that $\eta (s)$ has no singularity, and note that, let's $z = \sigma + it$:
$$\eta (s ) = \sum_{n \geq 0} \frac{(-1)^{n-1}}{n^s} =\sum_{n \geq 0}\frac{(-1)^{n-1}\cos (t\log n)}{n^\sigma}-i\sum_{n \geq 0}\frac{(-1)^{n-1}\sin(t\log n)}{n^\sigma}$$
By Abel's test, one obtains that $\eta(s)$ converges or bounded for all $z$ in $V$.
Also by writing:
$$\eta (s ) = \sum_{n \geq 1} \frac{1}{(2n-1)^s} – \frac{1}{(2n)^s} = \sum_{n\geq 1} f_n(s), s \in V$$
So $f_n(s)$ is holomorphic for all $s$ in $V$ and by the inequality:
$$\left\vert \frac{1}{(2n-1)^s} – \frac{1}{(2n)^s} \right\vert \leq \left\vert\frac{s}{(2n-1)^{s+1}} \right\vert, \Re{s} \geq 0$$
implies the uniform convergence of the sum of the series. This, therefore, shows that $\eta (s)$ is holomorphic and bounded in $V\setminus \{z_0\}$, and hence, $z_0$ is a removable singularity.
My question is whether what I did is correct ? Anything that needs to be added? Thank you.
Update:
There is an obnoxious error in my solution which is that I missed the roots of $1-2^{1-s}$. However, this helps me discover another way around:
Let $S_n (s)$ and $T_n$ be the partial sum of $\eta (s)$ and $\zeta (s)$, meaing:
$$S_{2n}(s) = \sum_{k\geq 1}^{2n} \frac{(-1)^k}{k^s} = (1-2^{1-s})T_{2n} + 2^{1-s} \sum_{n \geq k \geq 1} \frac{1}{(n+k)^s} $$$$=
(1-2^{1-s})T_{2n} +(2/n)^{1-s}\cdot \frac{1}{n} \sum_{n \geq k \geq 1} \frac{1}{(1+k/n)^s}$$
So as $n \to \infty$:
$$ \eta\left(1 + \frac{2\pi i}{\log 2}\cdot k\right) = \lim_{n\to \infty}S_{2n}\left(\frac{2\pi i}{\log 2}\cdot k\right)$$$$ = 0 + 0\cdot \int_{0}^{1} \frac{1}{(1+x)^{1 + \frac{2\pi i}{\log 2}\cdot k}}dx= 0 , k \in \mathbb{Z}\setminus \{0\} $$
This implies $\zeta (s)$ is bounded for the root of $1-2^{1-s}$.
Best Answer
A different solution is to note that the OP proved that $\zeta$ can only have at most simple poles at the roots of $1-2^{1-s}$ on $\Re s =1$ which are the roots of $2^{it}=1$ and of course we know that there is one such for $t=0$ so we need to exclude the other solutions so the ones where $t\log 2=2k \pi, k \ne 0$
For this one can use a $3$ (or any other odd number) trick, namely considering a similar function like $\eta$ that gives a set of possible poles at roots of $1-3^{1-s}$ on the line $\Re s =1$; since those are roots of $t\log 3=2k \pi$ and it is well known and (easy to prove) that $\log 2/\log 3$ is irrational (otherwise $2^b=3^a$ for some integers $a,b$ of which $b \ne 0$ as a denominator etc). This immediately implies that $0$ is the only common root of the two equations above, hence proving the required result about zeta.
The function in cause is $\eta_3(s)=1+1/2^s-2/3^s+1/4^s+1/5^s-2/6^s+...$; a similar grouping as for $\eta$ shows (conditional) convergence up to $\Re s >0$ and another similar trick like for $\eta$ shows that $\eta_3(s)=(1-3^{1-s})\zeta(s)$ for $\Re s >1$ where the series converges absolutely, giving the continuation of $\zeta$ to $\Re s >0$ except for those possible poles at the zeroes of the denominator $1-3^{1-s}$ so we are done!