The notation I have does not fall with the notation in the wiki link and I want to see I understood.
Can I expand the following idicator function: $$ 1\{x\leq 0\}x$$ in the following way? $$1 = \begin{cases} x & \quad x\leq 0\\ 0 & \quad \text{else }\\ \end{cases} $$
Your question is somewhat confusing. Do you mean indicator function?
If so, then $1\{x\leq0\}x$ would be more conveniently written (if I understand what you mean) as $\mathbf{1}_{x\leq0}(x)\cdot x$; $\mathbf{1}$ to stress that this is not multiplication by unity, the subscript $_{x\leq0}$ again to stress that this is an indicator function, and evaluated at $(x)$ since the indicator function is a function.
Having said this, $$\mathbf{1}_{x\leq0}(x)=\begin{cases}1&\text{ if }x\leq0\\0&\text{ otherwise }\end{cases}$$ so that $\mathbf{1}_{x\leq0}(x)\cdot x=x$ if $x\leq0$ and $\mathbf{1}_{x\leq0}(x)\cdot x=0$ otherwise.
The notation you mention seems to be using the Iverson bracket notation: $$ [x\le0] = \begin{cases} 1, & \quad x\leq 0\\ 0, & \quad \text{otherwise }\\ \end{cases} $$ Then the function you want is $f(x) = [x\le0] x$.