Spectrum: Products Given a unital C*-algebra $1\in\mathcal{A}$.
For the spectrum of products:
$$\sigma(AB)\cup\{0\}=\sigma(BA)\cup\{0\}$$
How can I find an example that shows 
that zero must be indeed taken out?
Obviously, they must not commute. 
So the natural choice will be matrices. 
But all matrices I tried out didn't work. 
Also taking ladder operators seem to fail?
What would be the idea behind such a recipe? 
(Besides I'm not asking for a specific example.)
 A: Take $A=B(\ell^2(\mathbb N))$ and consider the unilateral shift $S$. It's an isometry, so $S^*S=I$ is invertible and $0$ is not in the spectrum. But $SS^*$ is not invertible (it's not surjective), so $0\in \sigma(SS^*)$
A: To add to Shirley's answer, you will not find such matrices (if you want them square, i.e. in a C$^*$-algebra). For square matrices $A,B$, the two matrices $AB$ and $BA $ have the same eigenvalues (counting multiplicities). If you allow rectangular matrices for the question, you can take 
$$
A=\begin{bmatrix}1\\0\end{bmatrix}, \ \ \ \ \ B=A^T. 
$$
Then $BA=1$, with spectrum $\{1\}$, and $AB=E_{11}$ has spectrum $\{0,1\}$. 

Here is a proof of the fact that $AB$ and $BA$ have the same eigenvalues (counting multiplicities) when both $A$ and $B$ are square. Let
$$
M=\begin{bmatrix} I&A\\ 0&I\end{bmatrix}.
$$
It is easy to see that $M$ is invertible and its inverse is $$
M^{-1}=\begin{bmatrix} I&-A\\ 0&I\end{bmatrix}.
$$
Then a straightforward computation shows that 
$$
M^{-1}\,\begin{bmatrix} AB&0\\ B&0\end{bmatrix}\,M
=\begin{bmatrix}0&0\\ B& BA\end{bmatrix}.
$$
So the two matrices above are similar, and in particular they have the same characteristic polynomial. They have, respectively, characteristic polynomials $$P_1(t)=t^n\,P_{AB}(t),\ \ \ \ P_2(t)=t^n\,P_{BA}(t).$$
Thus $P_{AB}(t)=P_{BA}(t)$ and so $AB$ and $BA$ have the same eigenvalues with the same multiplicities. 
