Is there some way to express:

$$I(t) = \int_{-\infty}^{t} e^{-2\mathrm{cosh}(x)}~\mathrm{d}x$$

From Bessel functions?

By substituting $y = \mathrm{cosh}(x)$ we get

$$I(t) = \int_{1}^{\mathrm{cosh}(t)} \frac{e^{-2 y}}{\sqrt{y^2-1}}~\mathrm{d}y$$

In this form, Mathematica will give $I(\infty) = \mathrm{BesseK}(0,2)$ but not indication of what $I(t)$ might be.

I can write the integral as:

$$I(t) = \int_{1}^{\mathrm{sinh}(t)} \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}~\mathrm{d}z$$

and try to expand $f(x) = \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}$ as

$$f(z) = \sum_{i=0}^{\infty} (-1)^j\frac{2~\mathrm{BesselK}\left(\frac{1}{2}+j,2\right)}{j!\sqrt{\pi}} z^{2j}$$

Unfortunately the series diverges for $z \ge 1$, as does the series for $f(z) z e^{2 z}$