Well, we are looking at the inverse Laplace transform of:
$$\text{y}_{\space\text{n}}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\cdot\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(t\right)}\tag1$$
Where $\mathcal{I}_{\space\text{p}}\left(\text{z}\right)$ is the modified Bessel function of the first kind.
Now, using the convolution theorem, of the Laplace transform, we can write:
$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag2$$
Using the table of selected Laplace transforms, we can write:
$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}=\delta\left(t-\tau-\text{n}\right)\tag3$$
Where $\delta\left(x\right)$ is the Dirac Delta function.
So, we can rewrite equation $\left(2\right)$ as follows:
$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau\tag4$$
Using the definition of the modified Bessel function of the first kind, we can write:
$$\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\tag5$$
Where $\Gamma\left(\text{s}\right)$ is the Gamma function.
So using the table of selected Laplace transforms, we can write:
$$\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\right]_{\left(\tau\right)}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\tag6$$
Now, we can rewrite equation $\left(4\right)$ as follows:
$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\right\}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau\tag7$$
When $\text{n}>0\space\wedge\space\text{k}\ge0\space\wedge\space t\ge0$, we can write:
$$\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau=\frac{2\cdot\theta\left(t\right)-1}{\left(t-\text{n}\right)^{1+2\text{k}}}\cdot\theta\left(t-\text{n}-t\cdot\theta\left(-t\right)\right)\cdot\theta\left(\text{n}-t+t\cdot\theta\left(t\right)\right)\tag8$$
Where $\theta\left(\text{a}\right)$ is the Heaviside step function.