Let $u=x+A/x$ so that $x^2-u x+A=0$, or
$$x = \frac{u}{2} \pm \frac12 \sqrt{u^2-4 A} $$
$$dx = \frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 A}} \right ) du$$
Note that the integration limits provided by the mapping $x \mapsto u$ depend on whether the point $x=B$ is less than or greater than the minimum of $u$ at $x=\sqrt{A}$. Let's assume the former, i.e., $B \lt \sqrt{A}$; then the integral over $u$ is
$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 - \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u}\\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \sqrt{A}\int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} \end{align} $$
where
$$\alpha = \operatorname{arccosh}{\left (\frac{B+\frac{A}{B}}{2 \sqrt{A}} \right )}$$
If however $B \ge \sqrt{A}$,
$$\begin{align} I(A,B) &= \frac12 \int_{B+A/B}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 \int_{B+A/B}^{2 \sqrt{A}} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} + \frac12 \int_{2 \sqrt{A}}^{\infty} du \, \left (1 + \frac{u}{\sqrt{u^2-4 A}} \right ) e^{-u} \\ &= \frac12 e^{-\left (B+\frac{A}{B} \right )} + \sqrt{A} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}}\end{align} $$
For either case, let's consider
$$\begin{align} \int_{\alpha}^{\infty} dt \, \cosh{t} \; e^{-2 \sqrt{A} \cosh{t}} &= \int_{\alpha}^{\infty} dt \, \coth{t} \sinh{t} \; e^{-2 \sqrt{A} \cosh{t}} \\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{2 \sqrt{A}}\int_{\alpha}^{\infty} dt \, \frac1{\sinh^2{t}} e^{-2 \sqrt{A} \cosh{t}}\\ &= \frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} - \frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} + \frac{3}{4 A} \int_{\alpha}^{\infty} dt \, \frac{\cosh{t}}{\sinh^4{t}} e^{-2 \sqrt{A} \cosh{t}} \end{align} $$
We could go on and develop more terms in the series by further integration by parts, but let's take stock. Recall that
$$\cosh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} $$
$$\sinh{\alpha} = \frac{B+\frac{A}{B}}{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{1/2} $$
Then
$$\frac1{2 \sqrt{A}} \coth{\alpha}\; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac1{2 \sqrt{A}} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-1/2} e^{-\left ( B+\frac{A}{B} \right )} $$
$$\frac1{4 A} \frac1{\sinh^3{\alpha}} \; e^{-2 \sqrt{A} \cosh{\alpha}} = \frac{2 \sqrt{A}}{\left ( B+\frac{A}{B} \right )^3} \left [1-\frac{4 A}{\left (B+\frac{A}{B} \right )^2} \right ]^{-3/2} e^{-\left ( B+\frac{A}{B} \right )} $$
Note that additional terms will produce smaller contributions, so we can stop here for the moment and appreciate two uses of this series:
1. Perturbation for $0 \lt B \lt \sqrt{A}$ from the exact Bessel function result at $B=0$:
Using the assumption that $B \gt 0$ is a small perturbation, we expand out to $O(B^3)$, requiring the first two terms of the above expansion. Any further terms requires more integration by parts. Note that, in this region, it is the $1/B$ term that dominates. We find that
$$I(A,B) = 2 \sqrt{A} K_1 \left (2 \sqrt{A} \right ) - \left [ \frac1{2 A} B^2 + \frac1{2 A^2} B^3 +O(B^4)\right ] e^{-B-\frac{A}{B}}$$
Note that the perturbation term for small $B$ is extremely small. This has been verified numerically, although I caution that the extreme smallness of the perturbation makes it difficult to verify the exact expansion terms in $B$.
2. Global approximation of $I(A,B)$ for all $B$
We may accomplish this by considering expansions for both small $B$ and large $B$. Note that we use either one of the two cases delineated above according to whether $B$ is less than or greater than $\sqrt{A}$. I will not push this further, and I anticipate some difficulties because of the exponential term in the expansion, but I believe a two-point Pade approximant may do the trick.
