This proof is my understanding of a very interesting comment by @leoli1 on my previous related question about the following extended Riemann Zeta function which converges for $\sigma>0$ where $s=\sigma+it$.
$$\boxed{\zeta(s)=\frac{1}{1-2^{1-s}}\sum\frac{(-1)^{n+1}}{n^{s}}}$$
Question: Is my proof a correct understanding of @leoli1's suggestion?
Step 1
We find yet another series for $\zeta(s)$ using the following series:
$$ X(s)= \frac{1}{1^s} + \frac{1}{2^s} - \frac{2}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} - \frac{2}{6^s} + \ldots$$
Then
$$\begin{align} \zeta(s) - X(s) &= \frac{3}{3^s} + \frac{3}{6^s} + \frac{3}{6^s} + \ldots \\ \\ &= \frac{3}{3^s} \zeta(s) \end{align} $$
Rearranging gives us
$$ \boxed{\zeta(s) = \frac{1}{1-3^{1-s}} X(s) }$$
Step 2
We observe that $\sum(-1)^{n+1}/n^{s}$ is a Dirichlet series that converges for $\sigma>0$. Therefore it has no poles $\sigma>0$.
Similarly we observe that $X(s)$ is a Dirichlet series that converges for $\sigma>0$, because the sum of the coefficients $1+1-2+1+1-2 + \ldots$ is bounded. Therefore it has no poles $\sigma>0$.
This means any poles must come from the factors $(1-2^{1-s})^{-1}$ and $(1-3^{1-s})^{-1}$.
Step 3
The factors have poles at $s$ where
$$ 1- s = 2i\pi n \;/ \;\ln(2)$$
and
$$ 1-s = 2i\pi m \;/ \;\ln(3) $$
Where $n$ and $m$ are integers. However
$$ \frac{n}{m} = \frac{\ln(2)}{\ln(3)} $$
The RHS is irrational, and therefore only $n=m=0$ satisfies the expression.
Step 4
Therefore $\zeta(s)$ has only one pole, at $s=1$, corresponding to $n=m=0$.
Remark: If correct, this seems a very powerful way of showing the zeros of the sums $\sum(-1)^{n+1}/n^{s}$ and $X(s)$ cancel exactly the poles of the factors $(1-2^{1-s})^{-1}$ and $(1-3^{1-s})^{-1}$. Is this correct?