Question: How do I construct matrices $P\in\mathbb{C}^{\ n\ x\ n}$ such that:

1. They **are not** projectors (projector: $A=A^2$)
2. $Col(P)\cap Ker(P) = \{0\}$
3. $\forall x \in \mathbb{C}^n,\ x = y+z\quad \left(y\in Col(P),\ z\in Ker(P)\right)$

Answer: 
What you need is that your $P$ has no nilpotent part. That is, its Jordan form should be diagonal. In terms of blocks, your $P$ will be of the form $SXS^{-1}$ with $S$ invertible and 
$$
X=\begin{bmatrix}Y&0\\0&0\end{bmatrix},
$$
with $Y\in\mathbb C^{m\times m}$, $m\leq n$. 

At its simplest form, for instance, here is an example of such $P$:
$$
P=\begin{bmatrix}1&0&0\\ 0&2&0\\0&0&0\end{bmatrix}. 
$$
The column space is the span of the first two vectors in the canonical basis, while the kernel is the span of the third vector in the canonical basis.
It was easier than I thought. 

For example $T= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ has $T^2= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$, so $T\neq T^2$.
The $col(T)=\mathbb{C}^2$, and $ker(T)=0$. Then properties 2 and 3 are satisfied.

Another example $S= \begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix}$ has $S^2= \begin{bmatrix}4 & 0 \\ 0 & 0\end{bmatrix}$, so $S\neq S^2$.
The $col(S)=\begin{bmatrix}1\\ 0 \end{bmatrix}$, and $ker(T)=\begin{bmatrix} 0 \\ 1\end{bmatrix}$; so properties 2 and 3 are also satisfied.