As you mentioned, we know the Abramovich-Stegun approximation for the CDF
$$\Phi(z) = h(z) − \dfrac{\varphi(z)}{z} \left[ 1 - \dfrac{1}{z^2} + \mathcal{O}\left(z^{-4} \right)\right], \quad \text{for} \; |z| \to \infty.$$
Moreover, note that we can develop the term
$$
e^{\pm x/2} \varphi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right] \right) = \dfrac{1}{\sqrt{2\pi}}\exp \left\lbrace \pm \dfrac{x}{2} - \dfrac{1}{2}\left(\dfrac{x^2}{\sigma^2} + \dfrac{\sigma^2}{4}\right) \mp \dfrac{x}{2}\right\rbrace = \sqrt{2\pi} \varphi \left( \dfrac{x}{\sigma}\right) \varphi \left(\dfrac{\sigma}{2} \right),
$$
where $ \sqrt{2\pi} \varphi \left( \dfrac{x}{\sigma}\right) \to 1 $ for $\sigma \to \infty$ and $ \sqrt{2\pi} \varphi \left( \dfrac{\sigma}{2}\right) \to 1 $ for $\sigma \to 0$.
I will split the derivation in the two cases, $\sigma \to 0$ and $\sigma \to \infty$.
First case: $\sigma \to 0$
In the $\sigma\to0$ case this means
$$
e^{\pm x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]\right) \simeq e^{\pm x/2} \left\lbrace h(\theta \cdot x) - \dfrac{\varphi(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right])}{\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]} + \mathcal{O}\right\rbrace.
$$
Then, in the expression for $b$, we have
$$
\begin{aligned}
b & = \theta \left\lbrace e^{x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} + \dfrac{\sigma}{2} \right]\right) − e^{−x/2} \Phi \left(\theta \left[\dfrac{x}{\sigma} - \dfrac{\sigma}{2} \right]\right)\right\rbrace \\
& \simeq \theta \cdot h(\theta \cdot x) \left( e^{x/2} − e^{−x/2} \right) + \theta \cdot \varphi(x/\sigma) \left[ \dfrac{-\theta}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{\theta}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right]
\end{aligned}
$$
and doing simple algebra you can see that the last term in brackets equals (in the $\sigma \to 0$ limit)
$$
\left[ \dfrac{-1}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{1}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] = \dfrac{x}{\left( x / \sigma \right)^3}.
$$
Therefore, we have
$$ b \simeq \theta \cdot h(\theta \cdot x) \left( e^{x/2} − e^{−x/2} \right) + \varphi(x/\sigma) \cdot x \cdot \left(\sigma / x\right)^3.
$$
Second case: $\sigma \to \infty$
In the $\sigma\to\infty$ case this means
$$
e^{\pm x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]\right) \simeq e^{\pm x/2} \left\lbrace h(\pm \theta) - \dfrac{\varphi(\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right])}{\theta \left[\dfrac{x}{\sigma} \pm \dfrac{\sigma}{2} \right]}+ \mathcal{O} \right\rbrace.
$$
Then, in the expression for $b$, we have
$$
\begin{aligned}
b & = \theta \left\lbrace e^{x/2}\Phi \left(\theta \left[\dfrac{x}{\sigma} + \dfrac{\sigma}{2} \right]\right) − e^{−x/2} \Phi \left(\theta \left[\dfrac{x}{\sigma} - \dfrac{\sigma}{2} \right]\right)\right\rbrace \\
& \simeq \theta \left( e^{x/2} h(\theta \cdot x) − h(-\theta \cdot x) e^{−x/2} \right) + \theta \cdot \varphi(\sigma / 2) \left[ \dfrac{-\theta}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{\theta}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right]
\end{aligned}
$$
similarly to the previous case, now in the $\sigma \to \infty$ limit, we have
$$
\left[ \dfrac{-1}{\dfrac{x}{\sigma} + \dfrac{\sigma}{2}} + \dfrac{1}{\dfrac{x}{\sigma} - \dfrac{\sigma}{2}} \right] = - 4 / \sigma.
$$
Therefore, we have
$$ b \simeq e^{\theta \cdot x/2} - 4 / \sigma \cdot \varphi(\sigma/2),
$$
where I have used the definition of the Heaviside to cancel one of the two terms with $h(\cdot)$.
Hope this helps!
