Solution to the "near-Gaussian" integral $\int_{0}^{\infty} e^{- \Lambda \sqrt{(z^2+a)^2+b^2}}\mathrm{d}z$

Would a double expansion, once for the square root and once for the exponential, work? It seems like you're interested in the $b\rightarrow 0$ behavior so I tried it and the answer is of the form: $I=\sqrt{\frac{\pi}{4\Lambda}}e^{-\Lambda a}-\frac{b^{2}\pi}{4\sqrt{a}} erf(\Lambda a)+ O(b^{6})$ (The 4th order cancels). Cheers.

(too long for a comment)

I am rather sure that the integral as it stands has no closed form in itself; however, one might be able to derive a series with elliptic integral terms (you do have the square root of a quartic, after all) that hopefully quickly converges (and computing an elliptic integral is quite easy with the AGM).

I also note that if you express your quartic's coefficients wholly in terms of $s$ and $x$, your quartic factors into two quadratics with complex conjugate roots (due to the constraints you put on those two parameters):

$$(z^2+a)^2+b^2=(z^2-2xz+s^2+x^2)(z^2+2xz+s^2+x^2)$$

and letting $\mu=x+is$, the quartic can also be expressed as $(z^2-2\Re\mu z+|\mu|^2)(z^2+2\Re\mu z+|\mu|^2)=(z-\mu)(z-\bar{\mu})(z+\mu)(z+\bar{\mu})$, such that the only parameters you have to contend with in your integral are $\mu$ and $\Lambda$.

I'll edit this once I figure out how to derive the eliiptic integral series...