In general, you have the right approach, but there is a little mistake in your final expression. In fact, it is more complicated than what you have.
$F_Y(y)=Pr\left\{Y
Now, if $a$ and $b$ are positive, then you can rewrite it as you did (you just forgot to include $y$ in your expression):
$=\int_0^\infty \int_0^\infty Pr \left\{ aX_1
$=\frac{1}{\Omega_2}\frac{1}{\Omega_3}\int_0^\infty \int_0^\infty (1-e^{-\frac{yx_2}{a\Omega_1}(1+bx_3)})e^{-\frac{x_2}{\Omega_2}}e^{-\frac{x_3}{\Omega_3}}dx_2dx_3$
$=\int_0^\infty \int_0^\infty (1-e^{-\frac{y\Omega_2z_2}{a\Omega_1}(1+b\Omega_3z_3)})e^{-z_2-z_3}dz_2dz_3$
and you can easily calculate it till the final closed form expression.
However, if $a$ is negative, then you must have switched the inequality:
$Pr \left\{ aX_1\frac{y}{a}x_2(1+bx_3) \right\}=e^{-\frac{yx_2}{a\Omega_1}(1+bx_3)}$ for negative $y$.
If $b$ is negative, then you must split the initial integral into parts when $(1+bx_3)$ is positive ($x_3<-1/b$), and when it is negative (x_3>-1/b).