Let \(\displaystyle{x}^{{\ast}}{\left({s}\right)}={\left({\frac{{{1}}}{{{\left({s}+\mu_{{1}}+\mu_{{2}}\right)}{\left({s}+\hat{{\lambda}}_{{2}}\right)}{\left({s}+\lambda_{{1}}+\lambda_{{2}}\right)}}}}\right)}\) be the laplace transform in question, where

You right, first thing here is a convolution
$x\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{(s+{\mu }_1+{\mu }_2)(s+{\hat{\lambda }}_2)(s+{\lambda }_1+{\lambda }_2)}\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_1+{\mu }_2}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\hat{\lambda }}_2}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_1+{\lambda }_2}\right)=f\left(t\right)*g\left(t\right)*w\left(t\right)$
where ${\displaystyle f\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_1+{\mu }_2}\right)}$ , ${\displaystyle g\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\hat{\lambda }}_2}\right)}$ and ${\displaystyle w\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_1+{\lambda }_2}\right)}$
Just to note, since convolution is associative and commutative, you can evaluate it in any order (which is why no brackets were needed above),
${\displaystyle x\left(t\right)=(f\left(t\right)\cdot g\left(t\right))\cdot w\left(t\right)=f\left(t\right)\cdot (g\left(t\right)\cdot w\left(t\right))=(g\left(t\right)\cdot f\left(t\right))\cdot w\left(t\right)=f\left(t\right)\cdot (w\left(t\right)\cdot g\left(t\right))=(f\left(t\right)\cdot w\left(t\right))\cdot g\left(t\right))}$
Now use the formula ${\displaystyle (f\cdot g)\left(t\right)={\int }_0^{t}f\left(\tau \right)g(t-\tau )d\tau }$ and finish the transform.