= Solution
If $4a^3+27b^2=0$ and $(a,b)\ne(0,0)$, there is a nonzero real number $r$ such that
$$
a=-3r^2,
\qquad b=2r^3.
$$
Then $(0,0,r)$ is singular and
$$
z^3+az+b=(z-r)^2(z+2r).
$$
For $r>0$, the singular point is locally isolated because all terms in
$$
x^2+y^2+(z-r)^2(z+2r)=0
$$
are nonnegative nearby. For $r<0$, the surface is locally a double cone with vertex at that point. Neither neighbourhood is homeomorphic to an open disc, so
$$
\boxed{S_{a,b}\text{ is not a smooth surface when }4a^3+27b^2=0.}
$$
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