Integrate $\frac{1}{x\,\log{x}}$ by parts

Your conclusion is not correct.

It only says that the two indefinite integrals can differ by $1$, which is a special case of the fact that antiderivatives of the same function can differ by a constant.

If you replace the undefinite integrals by definite ones, you end up with

$$I=0+I,$$ which is harmless. The devil is the integration constant.

You may rescue the reasoning by writing

$$I=\frac{\log x}{\log x}+C+I$$ and you conclude that $C=-1$ (and there remains another integration constant in $I$).