I think I have found all the possible solutions, but my method seems very clumsy. I would love to know if there are any methods that do not require so much checking.
I first defined an order $a\geq b \geq c \geq d.$ Then, I used the binomial expansion in order to factorise as follows: $$(a+b)^3=\left((-c )+(-d)\right)^3$$
$$a^3 +3a^2 b + 3ab^2 + b^3 = -c^3 - 3c^2 d - 3cd^2 - d^3$$
$$3(a^2 b + ab^2 + c^2 d + cd^2)=-24$$
$$ab(a+b) + cd(c+d)=-8$$
$$(ab-cd)(c+d)=8.$$
Thus, the factors on the left can be either $$ {\{1,8\}}, {\{-1,-8\}}, {\{2,4}\}, {\{-2,-4}\}. $$
By the order defined above, $c+d$ must always be negative, and so I manually tested the four cases: $$(i) c+d=-8,\quad (ii)c+d=-1,\quad (iii)c+d=-4,\quad (iv)c+d=-2.$$
For the cases $(ii)$ and $(iii)$ there were no integer solutions.
For the cases $(i)$ and $(iv)$ there was, respectively, one solution each. These are:
$$\{5,3,-4,-4\}, \quad \{3,-1,-1,-1\}.$$