if $a=b=0$
Then you have
$$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
This means that every basis $(v_1,v_2)$ of $\Bbb{R}^2$ is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of $A$ is $2$ and the fact that multiplying the zero matrix by $v_1$ and $v_2$ we get the vector $(0,0)^T$.
If $a=-b$ Then we have
$$(J-aI)=\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}=\begin{pmatrix}0&0\\0&-b+b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
You can do the same reasoning as above to conclude that every basis of $\Bbb{R}^2$ is a basis of eigenvectors.
If $a\neq -b$ we have:
$$(J-aI)v_1=0 \iff\begin{pmatrix}0&0\\0&-b-a\end{pmatrix}v_1=0\implies v_1 = k\cdot \begin{pmatrix}1\\0\end{pmatrix},\quad k\in \Bbb{R},$$
because $$\begin{pmatrix}0&0\\0&-b-a\end{pmatrix} \begin{pmatrix}k\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}.$$
Do you get the second eigenvector?