"Proof" that $\log2=0$ using the expansion of $\log(1+x)$

Well, there's one mistake along the way that I have spotted: $$\cdots=\left\{\left(1+\frac13+\frac15+\frac17\color{red}{+\cdots}\right)-\left(\frac12+\frac14+\frac16 +\frac18\color{red}{+\cdots} \right)\right\}$$

In both of the smaller parenthesis, you have a variation of the harmonic series, and neither $$1+\frac13+\frac15+\frac17\color{red}{+\cdots}$$ nor $$\left(\frac12+\frac14+\frac16 +\frac18\color{red}{+\cdots} \right)$$ converges. You cannot split an infinite sum to two inifinite sums if either of the three infinite sums does not converge.

You may look in Riemann series theorem The theorem of Riemann says that any conditionally convergent series (yours being the 'classic' example) may be rearranged to converge to any given number (including $\pm \infty$).

If  both series $$ \sum_{n=1}^{\infty} a_{2n-1} \qquad\text{and}\qquad \sum_{n=1}^{\infty} a_{2n} $$ are convergent, then also $$ \sum_{n=1}^{\infty}(-1)^{n+1}a_n $$ is convergent and $$ \sum_{n=1}^{\infty}(-1)^{n+1}a_n= \sum_{n=1}^{\infty} a_{2n-1} - \sum_{n=1}^{\infty} a_{2n} $$ (the proof is not difficult, change sign to the even numbered terms in order to simplify it and get rid of the $(-1)^{n+1}$). There's no requirement for absolute convergence here.

Unfortunately, in your case you cannot apply this result, because neither the odd-numbered term series nor the even-numbered term series converge.