As you point out, this is no harder than solving a general cubic. It turns out that it is also no easier.

For any choice of the $5$ parameters $a,b,c,d,e$ your desired equation can be multiplied out to a cubic $X^3+Ax^2+Bx+C$ which can be solved by Cardano's formula. You hope for a nicer solution perhaps based on the form. However one might expect that with $5$ parameters one can easily get $A,B,C$ to be anything desired. This turns out to be the case.

Here is one way assuming all parameters constrained to be real. The complex case is as easy.

Pick $a=b=0$ and $c=-(A+d)$ where $d$ is $+1$ or $-1$ according as $C \ge 0$ or $C \lt 0.$ Then $e$ and $f$ are the (real) roots $$\frac{Bd \pm \sqrt{B^2+4|C|}}{2}$$ of the quadratic equation. $$x^2-\frac{B}{d}x-\frac{C}{d}.$$

As you point out, this is no harder than solving a general cubic. It turns out that it is also no easier.

For any choice of the $6$ parameters $a,b,c,d,e,f$ your desired equation can be multiplied out to a cubic $x^3+Ax^2+Bx+C$ which can be solved by Cardano's formula. You hope for a nicer solution perhaps based on the form. However one might expect that with $6$ parameters one can easily get $A,B,C$ to be anything desired. This turns out to be the case.

Here is one way assuming all parameters constrained to be real. The complex case is as easy.

Pick $a=b=0$ and $c=-(A+d)$ where $d$ is $+1$ or $-1$ according as $C \ge 0$ or $C \lt 0.$ Then $e$ and $f$ are the (real) roots $$\frac{Bd \pm \sqrt{B^2+4|C|}}{2}$$ of the quadratic equation. $$x^2-\frac{B}{d}x-\frac{C}{d}.$$