Steepest descent of integrand with a movable saddle?
Let us here sketched a derivation.
OP's integral has a saddle point at $z=0$ with angular steepest descent direction $\frac{\pi}{4}$. Therefore, inspired by the method of steepest descent, we rewrite OP's integral as $$I(\lambda) ~:=~\int_{\mathbb{R}_+}\!\mathrm{d}z~ \exp\left\{-z^2+i\lambda\sqrt{1+z^2}\right\} ~=~ \frac{e^{i\lambda}}{2}J(\lambda),\qquad \lambda~>~0, \tag{1}$$ where $$\begin{align}J(\lambda) ~:=~&\int_{\mathbb{R}}\!\mathrm{d}z~ \exp\left\{-z^2+i\lambda\left(\sqrt{1+z^2}-1\right)\right\} \cr ~\stackrel{z=\frac{u}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right)}{=}& \frac{1}{\sqrt{\lambda}}\exp\left(\frac{i\pi}{4}\right) K(\lambda),\end{align}\tag{2} $$ where $$ K(\lambda)~:=~\int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-\frac{iu^2}{\lambda}-f_{\lambda}(u)\right\},\tag{3}$$ and where $$ -f_{\lambda}(u)~:=~i\lambda\left(\sqrt{1+\frac{iu^2}{\lambda}}-1\right) ~=~-\frac{u^2}{2} + \frac{iu^4}{8\lambda}+ O(u^6\lambda^{-2}).\tag{4} $$
Let us define the non-negative function $$ f(u)~:=~\frac{u^2}{2}. \tag{5} $$ Clearly $$ f_{\lambda}(u)~\longrightarrow~ f(u)\quad\text{for}\quad\lambda~\to~ \infty \tag{6}$$ $u$-pointwise.
We note that the real part is given by $$ {\rm Re}~ f_{\lambda}(u)~=~\lambda\sqrt{\frac{\sqrt{1+\frac{u^4}{\lambda^2}}-1}{2}} .\tag{7}$$ Define a non-negative function $$ g(u)~:=~ \frac{|u|}{2}~\theta(|u|\!-\!\sqrt{2}) , \tag{8}$$ where $\theta$ denotes the Heaviside step function. One may show that $$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\in\mathbb{R}:~~ {\rm Re}~ f_{\lambda}(u) ~\geq~ g(u),\tag{9}$$ because $$\forall\lambda\geq\frac{1}{\sqrt{2}}\forall u\geq\sqrt{2}:~~ \sqrt{1+\frac{u^4}{\lambda^2}}-1 ~\geq~ \frac{u^2}{2\lambda^2}.\tag{10}$$
It follows that $\exp\left\{-g(u)\right\}$ is a majorant function for the integrand (3). Lebesgue's dominated convergence theorem then shows that the corresponding integral (3) satisfies $$ K(\lambda)~\longrightarrow~ \int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-f(u)\right\}~=~\sqrt{2\pi} \quad \text{for} \quad\lambda~\to~ \infty. \tag{11} $$
Next we can estimate corrections to any order in $1/\lambda$ we like. E.g. $$\begin{align}K(\lambda)~\stackrel{(3)+(4)}{=}&\int_{\mathbb{R}}\!\mathrm{d}u~ \exp\left\{-\left(\frac{i}{\lambda}+\frac{1}{2}\right)u^2\right\} \left(1 +\frac{iu^4}{8\lambda} + O(\lambda^{-2})\right)\cr ~=~&\sqrt{\frac{\pi}{\frac{i}{\lambda}+\frac{1}{2}}}+\frac{i}{8\lambda}\underbrace{\int_{\mathbb{R}}\!\mathrm{d}u~\exp\left\{-\frac{u^2}{2}\right\}u^4}_{=3\sqrt{2\pi}} + O(\lambda^{-2})\cr ~=~&\sqrt{2\pi}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right) ,\end{align}\tag{12}$$ and so forth. Hence OP's integral (1) has an asymptotic expansion
$$ I(\lambda)~\sim~ e^{i\lambda}\frac{1+i}{2}\sqrt{\frac{\pi}{\lambda}}\left(1 -\frac{5i}{8\lambda} + O(\lambda^{-2})\right)\quad \text{for}\quad \lambda~\to~\infty .\tag{13} $$