The conjecture we are trying to prove is the following:
$$ x \in \mathbb{R^+} \implies -\sum_{n=1}^{\infty} \frac{\ln(x)^n}{(2^n-1)n!} = \sum_{n=1}^{\infty} \left( 1 - x^{\frac{1}{2^n}}\right)$$
Note the left hand side is absolutely convergent while the right hand side is conditionally convergent. I have verified these for about 12 decimal digits to be equal. Moreover, they both obey the same functional equation $f(x^2) = f(x) -x + 1$.
The proof of the functional equation for the right hand side is trivial so we show the left hand side
$$ -\sum_{n=1}^{\infty} \frac{\ln(x^2)^n}{(2^n-1)n!} = - \sum_{n=1}^{\infty} \frac{2^n\ln(x)^n}{(2^n-1)n!} = - \sum_{n=1}^{\infty} \left[\frac{\ln(x)^n}{(2^n-1)n!} + \frac{\ln(x)^n}{n!} \right] = $$ $$-\sum_{n=1}^{\infty}\frac{\ln(x)^n}{(2^n-1)n!} - \sum_{n=1}^{\infty} \frac{\ln(x)^n}{n!}= \sum_{n=1}^{\infty}\frac{\ln(x)^n}{(2^n-1)n!}-x+1$$
I have absolutely no reason to believe that just because these functions obey the same equation and are equal for at least 12 decimal digits (which is the limit of the calculator I'm using) that they are totally equal but this is a very odd coincidence to survive for so many digits.
They are in fact equal.
$$\sum_{n=1}^{\infty} \frac{\ln(x)^n}{(2^n-1)n!} = \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \frac{1}{1-2^{-n}}= \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2} \right)^n}{n!} \cdot \sum_{k=1}^\infty(2^{-n})^{k-1} $$ $$=\sum_{k=1}^\infty \sum_{n=1}^{\infty} \frac{\left(\frac{\ln(x)}{2^{k}} \right)^n}{n!} = \sum_{k=1}^\infty (e^{\frac{\ln(x)}{2^{k}}}-1)= \sum_{k=1}^\infty (x^{\frac{1}{2^k}}-1).$$