Because we cannot find, given a NTM $N$ and an input $x$, whether $N$ fulfills the condition $N(x)\not= x$ in any given finite amount of time. More precisely, we cannot enumerate all pairs $(N,x)$ such that $N(x)\not= x$, although we can enumerate all pairs $(N,x)$ such that $N$ halts on $x$ and $N(x)\not= x$.
Think for a moment. When can you consider $(N,x)$ fulfills the requirement $N(x)\not=x$?
- We run $N$ with input $x$ for some time and it stops. If it does not output $x$, great, we have found one.
- We run $N$ with input $x$. However $N$ does not stop yet. It seems $N$ just won't stop. We want very much to say $(N,x)$ does not fulfill that requirement; however, there is no moment we can be sure of that. $N$ is just running at all times. If we conclude $(N,x)$ does fulfill the requirement at some moment, it can happen $N$ will stop at the next moment, outputting $x$. Because of this possibility, we will never add $(N,x)$ when $N$ never stops on input $x$, even though $(N,x)$ does satisfy the requirement.
Note this is vastly different from the other case. When can we consider $(N,x)$ fulfils the requirement $N(x)=x$?
- We run $N$ with input $x$ for some time and it stops. If it does output $x$, great, we have found one.
- We run $N$ with input $x$. However $N$ does not stop yet. Fine, we just do not consider $(N,x)$ fulfils the requirement yet. Continue running $N$.
You might argue that NTM $N$ can guess $x$ such that $N(x)$ will halt and output $y\not=x$. So we will find one such $N$.
There is no problem in that.
However, consider the case when $N$ never stops on every input. Now, imagine the exact moment you consider that $N$ fulfills the condition that $N(x)\not= x$ for some $x$, using any algorithm. What is known to you at that moment? All you know is that $N$ has not stopped yet on any input. That, however, does not exclude the possibility that $N$ might stop later on for all inputs along all paths, always outputting the input. We can never exclude that possibility, forever. (More precisely, we can exclude that possibility for some $N$ for some inputs. However, there is no algorithm can determine that possibility for all $N$.)
If I chose a DTM in place of an NTM to run $M_w(x)$ for all possible inputs, it would be game over if $M_w(x)$ didn't halt because that would prevent subsequent runs of $M_w(x)$ with different $x$es.
It would also be game over if $M_w(x)$ will not halt for all $x$. Your $NTM$ will be running forever without ever outputting anything. That $(M,x)$ is in $L_2$ but your algorithm cannot claim it is in $L_2$ since there is no moment when the algorithm is able to claim that.
The explanation above is meant to be a way to understand the situation. A rigorous proof is still needed to confirm the conclusion mathematically (or "compute sciencely"). Since you have known a rigorous proof, I will not provide one here.