I would like to answer the question whether $$ (1+\frac{\log x}{x})^x\cdot\frac{\log x}{x}-1\sim \log x-1\text{ as }x\to\infty. $$
I think this boils down to the question whether $$ (1+\frac{\log x}{x})^x\sim x\text{ as }x\to\infty. $$
I think this is equivalent to $$ x\log(1+\frac{\log x}{x})\sim\log x\text{ as }x\to\infty $$
and this should be true since $\log x/x\to 0$ for $x\to\infty$, meaning that $$ \log(1+\frac{\log x}{x})\sim\frac{\log x}{x}\text{ as }x\to\infty, $$ hence $$ x\cdot\log(1+\frac{\log x}{x})\sim x\cdot \frac{\log x}{x}=\log x\text{ as }x\to\infty $$
Note that
$$ \left(1+\frac{\log x}{x}\right)^x=e^{x \log \left(1+\frac{\log x}{x}\right) }\sim e^{\log x}=x$$