Is the decomposition $$\text{Id}(z) = \chi_{\{ |u| \leq k\}}(z) + \chi_{\{|u| > k\}}(z)\tag{1}$$ well defined for $u \in L^p(0,T;L^q(\Omega))$? I guess (1) holds a.e.
So the problem is, is the set $$\{|u| \leq k\} = \{(x,t) : |u(x,t)| \leq k\}$$ well-defined?
I guess it makes sense (except we may miss a set of measure 0 in there) because $u$ is defined a.e. However I feel I am missing a justification.
For every function $u$, the set $\{u\le k\}$ is well-defined. But the elements of $L^p$ are not functions; they are equivalence classes of functions. Since you consider a Bochner space, let's recall what this means: $u\sim v$ if for almost every $t$, the equality $u(\cdot,t)=v(\cdot, t)$ holds for a.e. $x$.
Then $\{u\le k\}$ is really an equivalence class of sets, with two sets $A,B$ being equivalent if on almost every $t$-slice, their symmetric difference has zero measure in the $x$-space. By Fubini's theorem, this implies that $A\triangle B$ has zero measure in the $xt$-space.
But you may find it awkward to drag these equivalences through the computations. Instead, I suggest saying: pick a representative of $u$ (i.e., of its equivalence class of functions) and define $\{u\le k\}$ using this representative. Then you do whatever you wanted to do with this set, and show that in the end, the choice of the representative did not matter; you'd get the same thing.