= Solution
Under local integrability and suitable bounds away from zero, <Watson's lemma> gives
$$
\int_0^Ae^{-\lambda t}f(t)\,dt\sim\sum_{n\ge0}a_n\Gamma(\alpha+n\beta+1)\lambda^{-\alpha-n\beta-1},\qquad\lambda\to+\infty.
$$
Each term follows by $u=\lambda t$ and the <Gamma function> integral; a finite local remainder is bounded in the same way, and the region away from zero is exponentially small.
For <Laplace's method>, assume an isolated nondegenerate global minimum $t_*$ of $p$, with $p''(t_*)>0$ and suitable tail bounds. Expand $p,q$ near $t_*$ and set $t-t_*=v/\sqrt z$. The leading term is $e^{-zp(t_*)}q(t_*)\sqrt{2\pi/(zp''(t_*))}$; successive terms come from Gaussian moments of the Taylor expansions. Several equally low minima contribute additively, while degenerate minima need a different scale.
Here the integrand is entire in $t$, so deform the contour to the line at imaginary part $-\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the line at imaginary part $\pi$. The two horizontal tails sum to $2\int_0^\infty e^{-z\cosh x}\,dx=O(e^{-z}z^{-1/2})$. The vertical segment contributes $i\int_{-\pi}^{\pi}e^{z\cos\theta}\,d\theta$. Its maximum is at zero; putting $\theta=v/\sqrt z$ and using $\cos\theta=1-\theta^2/2+\theta^4/24+\cdots$ gives
$$
i e^zz^{-1/2}\int_{\mathbb R}e^{-v^2/2}\left(1+\frac{v^4}{24z}+O(z^{-2})\right)dv.
$$
The fourth Gaussian moment is three times the zeroth, so
$$
\boxed{\int_{-\infty-i\pi}^{\infty+i\pi}e^{z\cosh t}\,dt
=i\sqrt{2\pi}\,e^z\left(z^{-1/2}+\tfrac18z^{-3/2}+O(z^{-5/2})\right).}
$$
Since $\Gamma(1/2)=\sqrt\pi$, this has the required normalization.
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