Usually these require the Residue theorem. The inverse Laplace transform is given by
$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{\exp{(s t)}}{s (s^2 + \omega^2)} \frac{\cosh{(s \frac{x}{a})}}{\cosh{(s \frac{\ell}{a})}} $
where $c$ is a real number such that all of the singularities of the integrand lie to the left.
I will refer you to other posts to view a picture of the contour, which involves a semicircle along the line of integration oriented to the left, enclosing all of the poles of the integrand.
The poles of the integrand here are at $s=0$ and $s = \pm i \omega$. The residues corresponding to these poles (i.e., $\lim_{x \rightarrow p} (x-p) f(x)$ for integrand $f(x)$ and pole $p$) are as follows: $\frac{1}{\omega^2}$ and
$\frac{\exp{(\pm i \omega t)} \cos{(\frac{x}{a})}}{2 \omega^2 \cos{(\frac{\ell}{a})}}$
respectively. Adding the residues up, we get
$\frac{1}{\omega^2} \left( 1 - \frac{\cos{(\frac{x}{a})} \cos{(\omega t)}}{\cos{(\frac{\ell}{a})}} \right ) $
There are also poles resulting from the $\cosh{(s \frac{\ell}{a})}$ term in the integrand at $s=\pm i (2 k+1) (\pi/2) (a/\ell)$ for all integers $k$. The residues from these poles are
$\sum_{k=-\infty}^{\infty} (-1)^{k+1} \frac{\exp{(i (2 k+1) \frac{\pi}{2} \frac{a}{\ell} t)} \cos{((2 k+1) \frac{\pi}{2} \frac{x}{\ell} t)}}{i (2 k+1) \frac{\pi}{2} \frac{a}{\ell} \left ( \omega^2 - (2 k+1)^2 (\frac{\pi}{2} \frac{a}{\ell})^2 \right )} $
$= 2 \sum_{k=0}^{\infty} (-1)^{k+1} \frac{\sin{((2 k+1) \frac{\pi}{2} \frac{a}{\ell} t)} \cos{((2 k+1) \frac{\pi}{2} \frac{x}{\ell} t)}}{(2 k+1) \frac{\pi}{2} \frac{a}{\ell} \left ( \omega^2 - (2 k+1)^2 (\frac{\pi}{2} \frac{a}{\ell})^2 \right )} $
I do not know of a closed-form expression for this latter sum.