Asking if $ln(x)=y$ is just a fancy way of asking does $e^y=x$. We like to write things as $y=f(x)$. However, when we have $e^y=x$, we can't "solve" for $y$. Thus, the $ln(x)=y$ notation was formed to allow use to write $y$ as a function of $x$. If $ln(-1)=0$ is a correct statement, then $e^0=-1$. However, these is clearly false. Finally, lets try to do what Bernoulli did but writing in the original form not $ln$ form.
$0=ln(1)\implies 0=ln(-1*-1)\implies 0=ln(-1).$
However, what that translates to if we don't use the $ln$ shorthand is:
$1=e^0 \implies -1*-1=e^0, \implies -1=e*0$.
But if we look at the final implies that is not mathematically valid. We divided by $-1$ on the left side but not on the right side. Thus, it was not a valid operation when we did it in the $ln$ shorthand. Like the comments said, your bad assumption was assuming the laws worked for all numbers when they just work for positive ones. The reason why is what I laid out. The laws simply come from observing what happens when we are working with the exponentials not from what appears to work with the $ln$ shorthand.