The short answer is the usual one: if you try to apply an identity without paying attention to when that identity is valid, you're likely to get all sorts of nonsensical results.
It's fair to speculate that the identities may still hold in situations where you haven't checked... but as soon as you derive something suspicious, you should suspect your speculation was wrong.
That said, the equation you wrote is actually true in a certain sense: the complex logarithm, $\log$, is not a function: it's a "multi-valued function", and the set of values the left hand side can take is equal to the set of values the right hand side can take.
In particular, $\log(-1)$ has the values $(\pi + 2 \pi n) \mathbf{i}$, where $n$ ranges over all integers. It's easy to see that $\log(-1)$ and $-\log(-1)$ have the same set of values.
However, multi-valued functions are one of those things you should avoid like the plague until you actually learn them properly: it's way too easy to make mistakes. It's more "common to use the principal branch" of the complex logarithm, $\def\Log{\mathop{\mathrm{Log}}}\Log$, which really is an ordinary function.
But when you do, you have to pay attention to the fact that $\Log$ does not satisfy the identity $\Log(xy) = \Log(x) + \Log(y)$. However, we do know the difference between the left and right hand sides is either $0$, $2 \pi \mathbf{i}$, or $-2 \pi \mathbf{i}$. It's generally pretty easy to figure out which, once you understand what $\Log$ is computing.
As an aside, there are other variations on the logarithm where $\log(-1) = 0$. For these variations, the equation you wrote holds. :)