Not sure this goes all the way, but a start:
S = Solve[σ^3 + 3 r σ + s == 0, σ];
Use FullSimplify with specific transformation rules:
Clear[Jplus, Jminus]; (* delete your definitions *)
FullSimplify[S,
TransformationFunctions -> {Automatic,
Function[X, X /. Sqrt[4 r^3 + s^2] -> s - 2 Jminus],
Function[X, X /. Sqrt[4 r^3 + s^2] -> -s + 2 Jplus],
Function[X, X /. Jplus -> s - Jminus],
Function[X, X /. Jminus -> s - Jplus]}]
$$ \left\{\left\{\sigma \to \frac{J_- r}{(-J_-)^{4/3}}+(-J_-)^{1/3}\right\},\left\{\sigma \to \frac{i \left(\sqrt{3}+i\right) (-J_-)^{2/3}+i \sqrt{3} r+r}{2 (-J_-)^{1/3}}\right\},\left\{\sigma \to \frac{\left(-1-i \sqrt{3}\right) (-J_-)^{2/3}-i \sqrt{3} r+r}{2 (-J_-)^{1/3}}\right\}\right\} $$
Notice that I did the substitutions on the square roots, not on the entire $J_{\pm}$ expressions, for the reason @Bill mentions: pattern-matching is brittle and must be aided with such tricks.
Do you have any assumptions on the signs of $r$, $s$, $J_+$, and $J_-$? These could help to simplify further., for example:
S /. Sqrt[4 r^3 + s^2] -> s - 2 Jminus /. 1/(-Jminus)^(1/3) -> Jplus^(1/3)/r
$$ \left\{\left\{\sigma \to (-J_-)^{1/3}-(J_+)^{1/3}\right\},\left\{\sigma \to \frac{1}{2} \left(1+i \sqrt{3}\right) (J_+)^{1/3}-\frac{1}{2} \left(1-i \sqrt{3}\right) (-J_-)^{1/3}\right\},\left\{\sigma \to \frac{1}{2} \left(1-i \sqrt{3}\right) (J_+)^{1/3}-\frac{1}{2} \left(1+i \sqrt{3}\right) (-J_-)^{1/3}\right\}\right\} $$