= Solution
The exponent $\Phi(t)=\tfrac i2(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi(-i)=1$ and $\Phi(i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit <circle> and rotate $t=-i e^{i\theta}$. This gives the exact representation
$$
I_0(x)=\frac1{2\pi}\int_{-\pi}^{\pi}e^{x\cos\theta}\,d\theta.
$$
The dominant saddle at $\theta=0$ is a steepest-descent direction for the <method of steepest descent>: $\cos\theta=1-\theta^2/2+\theta^4/24-\cdots$. Set $\theta=s/\sqrt x$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.
To obtain every coefficient efficiently, the <integral> satisfies $x^2I_0''+xI_0'-x^2I_0=0$, by differentiating and integrating the <derivative> of $\sin\theta\,e^{x\cos\theta}$. Substitute $I_0=e^x x^{-1/2}\sum a_nx^{-n}/\sqrt{2\pi}$ to get $a_0=1$ and $a_n=(2n-1)^2a_{n-1}/(8n)$. Therefore the full <asymptotic expansion> is
$$
\boxed{I_0(x)\sim\frac{e^x}{\sqrt{2\pi x}}
\sum_{n=0}^\infty\frac{((2n-1)!!)^2}{n!\,8^n x^n}
=\frac{e^x}{\sqrt{2\pi x}}\left(1+\frac1{8x}+\frac9{128x^2}+\frac{225}{3072x^3}+\cdots\right).}
$$
Here $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to+\infty$. The infinite <series> is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their <Stokes phenomenon>.
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