It is indeed correct to say that
$$Y(j\omega)=\frac{1}{\alpha+j\omega}\cdot\frac{1}{\beta+j\omega}=\frac{1}{(\alpha+j\omega)(\beta+j\omega)}$$
You could write the answer just in that why. The problem with that is that it would be a bit hard to antitransform and come back to the time-domain to find $y(t)$. So let's manipulate that expression a little bit. Using the partial-fraction expansion we get:
$$Y(j\omega)=\frac{1}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}$$
where we want to calculate the constants $A$ and $B$. Rewriting those two fractions as one we get that:
$$\frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}=\frac{A(\beta+j\omega)+B(\alpha+j\omega)}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A\beta+jA\omega+B\alpha+jB\omega}{(\alpha+j\omega)(\beta+j\omega)}$$
So:
$$\frac{1}{(\alpha+j\omega)(\beta+j\omega)}=\frac{A\beta+jA\omega+B\alpha+jB\omega}{(\alpha+j\omega)(\beta+j\omega)}$$
That equality holds only if
$$1=A\beta+jA\omega+B\alpha+jB\omega=(A\beta+B\alpha)+j\omega(A+B)$$
As the left side of the equation is purely real, then the imaginary part of the right side must equal zero and we can calculate our constants as functions of $\alpha$ and $\beta$:
$$A=-B \implies 1 = A\beta - A\alpha \implies A =\frac{1}{\beta-\alpha} \implies B=-\frac{1}{\beta-\alpha}$$
So we can write the transform of the output as proposed in the beginning:
$$Y(j\omega)=
\frac{A}{\alpha+j\omega}+\frac{B}{\beta+j\omega}=
\frac{\frac{1}{\beta-\alpha}}{\alpha+j\omega}-\frac{\frac{1}{\beta-\alpha}}{\beta+j\omega}$$
Therefore our output is:
$$Y(j\omega) = \frac{1}{\beta-\alpha}\left(\frac{1}{\alpha+j\omega}-\frac{1}{\beta+j\omega}\right)$$
And now you only have to antitransform that (I believe you can do that) and that's your output in the time domain.
