Let $$ f(x)=1+\frac{\log x}{x}+o\left(\frac{\log x}{x}\right), x\to\infty. $$
Do we then have $$\lim_{x\to\infty}x\log(f(x))=\infty?$$
My answer would be yes, since:
First of all, I think that we have $$ f(x)\sim 1+\frac{\log x}{x}, x\to\infty~~~(1) $$ because $f(x)- (1+\frac{\log x}{x})=o\left(\frac{\log x}{x}\right)$ and for $g(x)=o\left(\frac{\log x}{x}\right)$ we have $g(x)\in o\left(1+\frac{\log x}{x}\right)$.
Thus, we have $$ \log(f(x))\sim\log\left(1+\frac{\log x}{x}\right)=\frac{\log x}{x}+O\left(\left(\frac{\log x}{x}\right)^2\right), x\to\infty $$
Consequently,
$$ x\log(f(x))\sim \log x+O\left(\frac{\log^2(x)}{x}\right), x\to\infty $$
and hence $$ \lim_{x\to\infty}x\log(f(x))=\lim_{x\to\infty}\left(\log x+O\left(\frac{\log^2(x)}{x}\right)\right)=\infty, $$ since each function $h(x)\in O\left(\frac{\log^2(x)}{x}\right)$ tends to $0$ as $x\to\infty$.
Would you agree?
We have that $${\log x \over x} \longrightarrow_{x\to \infty} 0$$ and $\log(1+y)\sim y$ as $y\to 0$. This means that $$\log(f(x)) = \log\left(1+{\log x\over x} + o\left({\log x\over x}\right)\right) \sim {\log x\over x}+o\left({\log x\over x}\right) \sim {\log x\over x}$$
Consequently $$x\log(f(x))\sim \log x \to \infty$$
Note that $1 + {\log x\over x} \sim 1$, so writing $f(x) \sim 1+ {\log x\over x} $ is not strong enough to get the result you want.