Why isn't $\ln z = \ln z + 2k\pi\mathrm{i}$ a contradiction? 
Define $$\operatorname{Ln} z = \operatorname{Ln} |z| + \mathrm{i}\operatorname{Arg}(z), \ \ \ \ \ z\in\Bbb{C}\backslash\{0\}, \ \ \ \ \ -\pi< \operatorname{Arg}(z) \leq \pi$$
  where $\operatorname{Ln}|z|$ is the usual real-valued natural logarithm. Furthermore, we state $$\ln z = \operatorname{Ln} |z| + \mathrm{i}\arg(z) = \operatorname{Ln} |z| + \mathrm{i}\left[\operatorname{Arg} (z) + 2k\pi\right], \ \ \ \ \ k\in\Bbb{Z}.$$ Distributing the imaginary unit $\mathrm{i}$ yields $$\ln z = \operatorname{Ln} z + 2k\pi\mathrm{i}, \ \ \ \ \ k\in \Bbb{Z}.$$ Exponential function is defined as $$\mathrm{e}^z = \sum_{n=0}^\infty\frac{z^n}{n!}.$$


From SCIPP's notes, in particular this handout on pages 6$-$7, it is seen that
$$\ln(\mathrm{e}^z) = z + 2k\pi\mathrm{i} \neq z \ \ \ \ \text{unless}\ \  k = 0.\tag1$$
This is in line with the definition as stated above. However, they go on to assert
$$\ln w = \ln w + 2k\pi\mathrm{i}, \ \ \ \ w \in \Bbb{C}\backslash\{0\}.\tag2$$


*

*Why is $(2)$ not a contradiction?


Assume instead $(2)$ is valid. Then substract $2k\pi\mathrm{i}$ from both sides of $(1)$ and $(2)$:
$$\ln(\mathrm{e}^z) -  2k\pi\mathrm{i} = z\tag{3}$$
$$\ln w - 2k\pi\mathrm{i} = \ln w.\tag{3'}$$
Take $w = e^z$ in $(3')$ and plug the result back into $(3)$. We arrive at
$$\ln(e^z) = z \overset{(1)}{\implies} z \neq z.$$
Where have I gone wrong?
 A: *

*In the world of real numbers, one always has
$$
e^{\ln x}=\ln(e^x)=x
$$
for $x>0$. However, one should not expect the same thing happens in the world of complex numbers. 

*(2) is not a contradiction by "definition" of the complex logarithm and the sentence "In fact, the latter is completely valid as a
set equality in light of eq. (43)." in the handout. It is very similar to saying that
$$
\int f(x)\ dx=\int f(x)\ dx+C\tag{*}
$$ 
which is indeed very confusing! One should really understand (using you notation in the post) $\ln z$ as a set, not a complex number. Thus one should avoid doing "cancellation" in (2) just as in (*) one could easily get the nonsense $0=1$ by doing cancellation (let $C=1$). 

*"Where have I gone wrong?"  The identity $\ln(e^z)=z$ and the very last "$\Rightarrow$" sign are incorrect. One needs to be careful that in (1), there is a phrase "unless $k=0$".
A: $$\ln w=\ln w+2k\pi i$$ is said nowhere to be an identity (which would make no sense).
Mind the nuances between $\ln z,\ln|z|,\text{Ln } z$ and $\text{Ln }|z|$.
