Writing $(1+s^2)^{-1/2} = s^{-1}(1+s^{-2})^{-1/2}$, expanding in a binomial series for $s>0$ and assuming that it is permissible to take the inverse transform term by term, show that the function $$Y(s) = c(1+s^2)^{-1/2}$$ has the inverse Laplace transform $$y(t)=c \sum_{n=0}^{\infty} \dfrac{(-1)^nt^{2n}}{2^{2n}(n!)^2}.$$
Firstly, I don't see how the rewriting is helpful. The series expansion for $(1+s^2)^{-1/2}$ around $s=0$ is given by $1+(-1/2)s^2+\dfrac{(-1/2)(-1/2-1)s^4}{2!}+...=\sum_{k=0}^{\infty}(-1)^k\dfrac{(2k)!}{2^{2k}(k!)^2}s^{2k}$
This is getting pretty close to the desired result. All I need to determine is $\mathcal{L^{-1}\{s^{2k}}\}$. How can I do this without turning to Mellin's formula?
Observe that you expanded $(1+s^2)^{-1/2}$ instead of $s^{-1}(1+s^{\color{red}{-}2})^{-1/2}.$ In particular, $$s^{-1}(1+s^{-2})^{-1/2} = s^{-1}\left(1-\frac{1}{2}s^{\color{red}{-}2}+\cdots\right).$$ Now use $$\mathcal{L}\left(t^n\right) = n! s^{\color{red}{-}(n+1)}$$ to compute the Laplace transform of $c(1+s^2)^{-1/2}$.