Q2 might still benefit from an explicit answer in terms of a special function.

The inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})=1/\cosh(\sqrt{2\lambda})$ follows from an entry in Table 2 in Theta functions; transform tables and examples for electrochemists:

$$\int_0^\infty dt\, e^{-\lambda t}\,\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2c}\biggl|\frac{i\pi t}{c^2}\right)=\frac{c\cosh(x\sqrt{\lambda})}{\cosh(c\sqrt{\lambda})},\;\;|x|<c, $$ where $\theta_1$ is a Jacobi theta function.
The inverse Laplace transform of $1/\cosh(\sqrt{2\lambda})$ is thus given by $${\cal L}^{-1}\left(\frac{1}{\cosh(\sqrt{2\lambda})}\right)=\frac{1}{\sqrt 2}\lim_{x\rightarrow 0}\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2\sqrt{2}}\biggl|\frac{i\pi t}{2}\right).$$

Q2 $\qquad$ "Is there a closed formula for the inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})$"
might still benefit from an explicit answer in terms of a special function.

The inverse Laplace transform of ${\rm sech}(\sqrt{2\lambda})=1/\cosh(\sqrt{2\lambda})$ follows from an entry in Table 2 in Theta functions; transform tables and examples for electrochemists:

$$\int_0^\infty dt\, e^{-\lambda t}\,\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2c}\biggl|\frac{i\pi t}{c^2}\right)=\frac{c\cosh(x\sqrt{\lambda})}{\cosh(c\sqrt{\lambda})},\;\;|x|<c, $$ where $\theta_1$ is a Jacobi theta function.
The inverse Laplace transform of $1/\cosh(\sqrt{2\lambda})$ is thus given by $${\cal L}^{-1}\left(\frac{1}{\cosh(\sqrt{2\lambda})}\right)=\frac{1}{\sqrt 2}\lim_{x\rightarrow 0}\frac{\partial}{\partial x}\theta_1\left(\frac{x}{2\sqrt{2}}\biggl|\frac{i\pi t}{2}\right).$$