A solution is given in the appendix of the following paper: Knessl, C. Exact and Asymptotic Solutions to a PDE That Arises in Time-Dependent Queues. Adv. Appl. Prob., 32(1) 256--283, 2000.
Let me describe this solution and highlight some subtle points.
${\rm Ai}(z)$ has different asymptotic approximations for large $|z|$ valid in different regions of the complex plane. We are going to use
$$
{\rm Ai}(z) \sim \frac{1}{2 \pi^{\frac{1}{2}} z^{\frac{1}{4}}}
\left( e^{-\frac{2}{3} z^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} z^{\frac{3}{2}}} \right) \;,
$$
which is valid as $|z| \to \infty$ assuming $\frac{\pi}{3} < {\rm arg}(z) < \frac{5 \pi}{3}$,
to replace
$$
f(z) = \frac{1}{{\rm Ai}^2(z)} \;,
$$
with
$$
g(z) = \frac{4 \pi z^{\frac{1}{2}}}
{\left( e^{-\frac{2}{3} z^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} z^{\frac{3}{2}}} \right)^2} \;.
$$
We need to consider the following three contours (with which we will take $R \to \infty$):
\begin{eqnarray}
C_1 &=& \{ {\rm i} s \}_{s = -R}^{s = R} \;, \\
C_2 &=& \{ R e^{{\rm i} \theta} \}_{\theta = \frac{\pi}{2}}^{\theta = \frac{3 \pi}{2}} \;, \\
C_3 &=& \{ e^{\frac{4 {\rm i} \pi}{3}} s \}_{s = R}^{s = 0} \cup
\{ e^{\frac{2 {\rm i} \pi}{3}} s \}_{s = 0}^{s = R} \;.
\end{eqnarray}
The given problem is to evaluate
$$
I \equiv \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1} f(z) \,dz \;.
$$
By Cauchy's residue theorem,
since the residues of the singularities of $f(z)$ are all zero, we have
$$
\lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1 \cup C_2} f(z) \,dz = 0 \;,
$$
thus
$$
I = - \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_2} f(z) \,dz \;.
$$
In view of the asymptotic approximation, it follows that
$$
I = - \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_2} g(z) \,dz \;.
$$
Again by Cauchy's residue theorem,
since the residues of the singularities of $g(z)$ are all zero, we have
$$
\lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1 \cup C_2} g(z) \,dz = 0 \;,
$$
thus
$$
I = \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_1} g(z) \,dz \;.
$$
Finally we may deform $C_1$ to $C_3$, i.e.,
$$
I = \lim_{R \to \infty} \frac{1}{2 \pi {\rm i}} \int_{C_3} g(z) \,dz \;.
$$
We can evaluate this integral by explicitly parametrizing $C_3$.
For the first part of $C_3$, $z = e^{\frac{4 \pi {\rm i}}{3}} s$,
where $s$ ranges from $\infty$ to $0$, and note that, in particular,
$z^{3/2} = s^{3/2}$.
For the second part of $C_3$, $z = e^{\frac{2 \pi {\rm i}}{3}} s$,
where $s$ ranges from $0$ to $\infty$, and here
$z^{3/2} = -s^{3/2}$. We have
$$
I = \frac{2}{{\rm i}} \int_\infty^0
\frac{s^{\frac{1}{2}}}
{\left( e^{-\frac{2}{3} s^{\frac{3}{2}}} + {\rm i} e^{\frac{2}{3} s^{\frac{3}{2}}} \right)^2} \,ds +
\frac{2}{{\rm i}} \int_0^\infty
\frac{-s^{\frac{1}{2}}}
{\left( e^{\frac{2}{3} s^{\frac{3}{2}}} + {\rm i} e^{-\frac{2}{3} s^{\frac{3}{2}}} \right)^2} \,ds \;.
$$
By multiplying and dividing each integral by the square of the complex conjugate of the term appearing in the denominator (the usual trick to make the denominator real-valued) and combining the integrals and simplifying we arrive at
\begin{eqnarray}
I &=& 8 \int_0^\infty \frac{s^{\frac{1}{2}}}
{\left( e^{\frac{4}{3} s^{\frac{3}{2}}} + e^{-\frac{4}{3} s^{\frac{3}{2}}} \right)^2} \,ds \\
&=& \int_0^\infty \frac{1}{{\rm cosh}^2(u)} \,du \\
&=& {\rm tanh}(u) \big|_0^\infty \\
&=& 1 \;,
\end{eqnarray}
as required.