It holds, but you have to be careful with complex logarithms.
Your mistake was in assuming that log 1 = 0. This is true when you're dealing with real numbers only, but not necessarily true when you're dealing with complex numbers.
The actual complex log of 1 is ##2k{\pi}i##, where k is an integer. This just becomes the familiar zero when k = 0. But k can be any integer, meaning the complex log has an infinite number of possible values.
More generally, the complex log of a complex no. z is given by ##\log(z) = \ln|z| + i(\theta + 2k\pi)##, where ##\theta## is the argument of the complex no. ##(-\pi < arg(z) < \pi)## and ##\ln|z|## is the usual single-valued real logarithm of the modulus of z.
The principal value of the complex log is often denoted as ##Log(z)## (the capitalisation is intended). It is defined by ##Log(z) = \ln|z| + i\theta##, where the argument ##\theta## lies in the same range as previously defined.
On that basis you can say that your statement holds insofar as the sum of the principal complex logs on the LHS will equal to one possible value of the multivalued complex log on the RHS.
As an example, working with the principal values of the complex logs, ##\log(-1) = i\pi##, so ##2\log(-1) = 2i\pi##, which is also one of the complex values of ##\log 1##.
Note that it is not necessarily the case that the sum of the principal logs on the LHS will equal to the principal log of the RHS. The question you posed is a counterexample, because the principal value of the log of 1 is the usual value, 0.
