$\newcommand{\R}{\mathbb R}\newcommand\sgn{\operatorname{sgn}}\newcommand{\vpi}{\varphi}$Obviously, for $a:=\sqrt{2\pi}\,\psi(0)$ we have
\begin{equation*}
\psi(0)=f(0),
\end{equation*}
where
\begin{equation*}
f:=a\vpi
\end{equation*}
and $\vpi$ is the standard normal density. Letting now $g(x):=\frac{\psi(x)-f(x)}x$ for $x\ne0$, with $g(0):=\psi'(0)$, we see that $g$ is a smooth integrable function and
\begin{equation*}
\psi(x)=f(x)+xg(x) \tag{1}\label{1}
\end{equation*}
for all real $x$.
So,
\begin{equation*}
I(t)=\frac{J_f(t)+J_h(t)}{\sqrt t}, \tag{2}\label{2}
\end{equation*}
where $h(x):=xg(x)$,
\begin{equation*}
J_f(t):=
\int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)}
\int_{\R}dx\,f(x)e^{ix^2y^2},
\end{equation*}
\begin{equation*}
z_1:=e^{i\theta_1},\quad z_2:=e^{i\theta_2},
\end{equation*}
with $J_h(t)$ defined similarly. Here and in what follows, $t$ is any real number $>0$, unless specified otherwise.
Note that
\begin{equation*}
\Im z_1\ne0\quad\text{and}\quad\Im z_2\ne0. \tag{3}\label{3}
\end{equation*}
Note also that
\begin{equation*}
\int_{\R}dx\,f(x)e^{ix^2y^2}=\frac a{\sqrt{1-2 i y^2}}
\end{equation*}
and hence
\begin{equation*}
\begin{aligned}
J_f(t)&=
a\int_0^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}.
\end{aligned}
\tag{4}\label{4}
\end{equation*}
For any fixed real $A>0$, by dominated convergence (which holds in view of \eqref{3}),
\begin{equation*}
\int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}
\to\frac1{z_1z_2}\int_0^A\frac{dy}{\sqrt{1-2 i y^2}}\ll1
\end{equation*}(as $t\to\infty$).
We write $E\ll F$ to mean $|E|=O(F)$.
So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$ and
\begin{equation*}
\int_0^A\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}}
=o(\ln t).
\end{equation*}
Also, we will have
\begin{equation*}
\int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)\sqrt{1-2 i y^2}} \\
\sim\int_A^{\sqrt t}\frac{dy}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)y\sqrt{-2 i}}
\sim\frac{1+i}{4z_1z_2}\,\ln t. \tag{5}\label{5}
\end{equation*}
The latter asymptotic expression in \eqref{5} can be obtained by taking the latter integral in \eqref{5} in closed form, by partial fraction decomposition. It can also be obtained by writing $\int_A^{\sqrt t}=\int_A^{t^b}+\int_{t^b}^{t^{1/2}}$ for $b\in(0,1/2)$ such that $b$ is close to $1/2$.
So, assuming $\psi(0)\ne0$, by \eqref{4},
\begin{equation*}
J_f(t)\sim
a\frac{1+i}{4z_1z_2}\,\ln t
=\frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\ln t.
\tag{6}\label{6}
\end{equation*}
Next,
\begin{equation*}
J_h(t)= \int_{\R}dx\,\sgn(x)g(x)K(t,x), \tag{7}\label{7}
\end{equation*}
where
\begin{equation*}
K(t,x):=\int_0^{\sqrt t}\frac{dy\,e^{ix^2y^2}|x|}{(\frac y{\sqrt t}-z_1)(\frac y{\sqrt t}-z_2)}
=\frac12\int_0^{r^2}dv\, e^{iv}H_r(v),
\end{equation*}
\begin{equation*}
r:=|x|\sqrt t,
\end{equation*}
\begin{equation*}
H_r(v):=\frac1{\sqrt v\,(\frac{\sqrt v}r-z_1)(\frac{\sqrt v}r-z_2)}.
\end{equation*}
Further,
\begin{equation*}
2K(t,x)=K_1+K_2, \tag{8}\label{8}
\end{equation*}
where
\begin{equation*}
K_1:=\int_0^{1\wedge r^2}dv\, e^{iv}H_r(v),\quad K_2:=\int_{1\wedge r^2}^{r^2}dv\, e^{iv}H_r(v),
\end{equation*}
$u\wedge w:=\min(u,w)$.
For $v\in(0,1\wedge r^2)$, we have $H_r(v)\ll\frac1{\sqrt v}$ and hence
\begin{equation*}
K_1\ll1. \tag{9}\label{9}
\end{equation*}
Note that $K_2=0$ if $r\le1$. If now $r>1$, then
\begin{equation*}
H'_r(v)=-\frac{H_r(v)}{2v}\,\Big(1+\frac1{\sqrt v-rz_1}+\frac1{\sqrt v-rz_2}\Big)
\ll \frac{|H_r(v)|}v\ll\frac1{v^{3/2}}
\end{equation*}
for $v>0$; so, integrating by parts, we see that
\begin{equation*}
K_2\ll1. \tag{10}\label{10}
\end{equation*}
Collecting \eqref{7}--\eqref{10} and recalling that $g$ is an integrable function, we see that
\begin{equation*}
J_h(t)\ll1. \tag{11}\label{11}
\end{equation*}
Collecting \eqref{2}, \eqref{6}, and \eqref{10}, we conclude that
\begin{equation*}
I(t)\sim
\frac{1+i}{4z_1z_2}\,\sqrt{2\pi}\,\psi(0)\frac{\ln t}{\sqrt t}, \tag{12}\label{12}
\end{equation*}
as $t\to\infty$.
As seen from the above proof, for \eqref{12} to hold it is enough that the function $\R\setminus\{0\}\ni x\mapsto\frac{\psi(x)-\psi(0)}x$ be integrable, with $\psi(0)\ne0$. (Of course, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
Concerning the case $\psi(0)=0$: Then the bounds on $H_r(v)$ and $H'_r(v)$ developed above provide for dominated convergence. So, taking into account that for each real $x\ne0$
\begin{equation}
H_r(v)\to\frac1{z_1z_2\sqrt v},
\end{equation}
we have
\begin{equation}
K(t,x)\to\frac1{2z_1z_2}\int_0^\infty\frac{dv\,e^{iv}}{\sqrt v}
=\frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2}
\end{equation}
and
\begin{equation*}
J_h(t)\to C_\psi:= c_\psi \frac{1+i}{2z_1z_2}\,\sqrt{\frac\pi2},
\end{equation*}
where
\begin{equation}
c_\psi:=\int_{\R}dx\,\sgn(x)g(x)=\int_{\R}dx\,\frac{\psi(x)}{|x|}.
\end{equation}
Therefore and because here $f=0$ and hence $J_f(t)=0$, we conclude that
\begin{equation*}
I(t)\sim \frac{C_\psi}{\sqrt t},
\end{equation*}
as $t\to\infty$, provided that $\psi(0)=0$, $c_\psi\ne0$, and the function $g$ is integrable. (In the case when $\psi(0)=0$, we have $g(x)=\psi(x)/x$ for $x\ne0$. As was already noted, $g$ will be integrable if, as stated by the OP, "$\psi$ is a smooth real-valued function with compact support".)
If $\psi(0)=0$ and $c_\psi=0$, then one has to dig deeper yet, possibly ad infinitum.
However, if $\psi$ is a "smooth bump", as you said in a comment, then apparently $\psi\ge0$ and hence $c_\psi>0$ (unless $\psi=0$ and hence $I(t)=0$).
We can evaluate $I(x)$ explicitly, and then asymptotically.
Indeed, using the substitution $s=ru/x$, we get
\begin{equation*}
I(x)=\frac1{\sqrt x}\lim_{R\to\infty}J_R(x), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
J_R(x)&:=\int_0^R dr\,e^{ir}\int_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}e^{-ru/x} \\
&=\int_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\int_0^R dr\,e^{(i-u/x)r} \\
&=\int_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac{1-e^{(i-u/x)R}}{u/x-i}.
\end{aligned}
\end{equation*}
Next, (i) for any real $u,x>0$ we have $\dfrac{1-e^{(i-u/x)R}}{u/x-i}\to\dfrac1{u/x-i}$ as $R\to\infty$, (ii) for any real $u,x,R>0$ we have $\Big|\dfrac{1-e^{(i-u/x)R}}{u/x-i}\Big|\le\dfrac2{|u/x-i|}\le2$, and (iii) for any real $x>0$ we have $\displaystyle{\int_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\,2<\infty}$.
So, by dominated convergence,
\begin{equation*}
\lim_{R\to\infty}J_R(x)=J(x), \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
&J(x):=\int_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac1{u/x-i} \\
&=\frac{x \left(-18 \ln x+\pi \left(8 i \sqrt{3} x+(9-9 i) \sqrt{2} \sqrt{x}-\frac{18
\sqrt[4]{-1}}{\sqrt{x}}+4 \sqrt{3}+9 i\right)\right)}{18 (-1+x (x-i))}
\end{aligned}
\tag{3}\label{3}
\end{equation*}
(note that the integrand in \eqref{3} is rational in $\sqrt u\,$, so that one can use partial fraction decomposition to get \eqref{3}).
So, as $x\to\infty$,
\begin{equation*}
J(x)\to c:=\frac{4 i \pi }{3 \sqrt{3}},
\end{equation*}
so that, by \eqref{1},
\begin{equation*}
I(x)\sim \frac c{\sqrt x}.
\end{equation*}
Best Answer
$$P(s)=2\int_{-\infty}^{\infty}{\rm d}p\int_{-\infty}^{\infty}{\rm d}q\ \delta\left(\frac{p^2+(p^2-q^2)^2}{p^2+4p^4}-s\right)e^{-\left(p^2+q^2\right)/a^2}$$ $$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}(x+4x^2)\int_{0}^{\infty}\frac{{\rm d}y}{\sqrt{y}}\ \delta\left(x+(x-y)^2-s(x+4x^2)\right)e^{-\left(x+y\right)/a^2}$$ $$\qquad=2\int_{0}^{\infty}\frac{{\rm d}x}{\sqrt{x}}e^{-x/a^2}\frac{x+4x^2}{z_+-z_-}\left[\frac{\theta(z_+)}{\sqrt{z_+}}e^{-z_+/a^2}+\frac{\theta(z_-)}{\sqrt{z_-}}e^{-z_-/a^2}\right]\theta\bigl(x-(1-s)/4s\bigr)$$ $$\qquad=\int_{(1-s)/4s}^{\infty}dx\,e^{-x/a^2}\frac{1+4x}{\sqrt{4 s x+s-1}}\left[\frac{1}{\sqrt{z_+}}e^{-z_+/a^2}+\frac{1}{\sqrt{z_-}}e^{-z_-/a^2}\right],$$ with $z_\pm=x\pm\sqrt{x (4 s x+s-1)}$ and $\theta(z)$ the unit step function. In the final equality I assumed $0<s<1/4$.
For $s\ll 1$ this approximates to $$P(s)\approx\frac{4}{\sqrt{s}}\int_{1/4s}^{\infty}dx\,e^{-2x/a^2}\frac{1}{\sqrt{4 s x-1}}=\sqrt{2\pi}(a/s)\exp\left(-\frac{1}{2a^2s}\right).$$