Does $\mathrm{\sum\limits_{-\infty}^\infty Ai(x)=1}$? Also on $\sum\limits_{-\infty}^0 \mathrm{Bi(x)}$. This question will be very similar to:

On $$\mathrm{\sum\limits_{x=1}^\infty Ci(x)}$$

and

On $$\mathrm{\sum_{x\in\Bbb Z}sech(x), \sum_{x=1}^\infty csch(x)}$$

all of which had closed forms, but it will use this Airy Ai function definition which is a type of Bessel function.
Note that I will actually focus on 2 constants which each converge slowly. I will use $\sum\limits_{n=a}^b A_n=\sum\limits_a^b A_n$ for shorthand and use the following notations. The reason these are split up is because one of the constants may possibly diverge. Note that $\mathrm A_{0,1}$ both have a lower bound of $0$, so the x=$0$ term will be subtracted:
$$\mathrm{A_0=\sum_{-\infty}^0 Ai(x), A_1=\sum_0^\infty Ai(x),A=A_0+A_1-Ai(0)=\sum_{-\infty}^\infty Ai(x)}$$
Here is a graph of the summand:

Here is a possible Abel-Plana formula computation.
$$\mathrm{A_0=\sum_0^{\infty}Ai(-x)\mathop=^{Abel}_{Plana}\frac12 Ai(-0)+\int_0^\infty Ai(-x) \,dx+\int_0^\infty\frac{Ai(- -ix)-Ai(- ix)}{e^{2\pi x}-1}\,dx=\frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac23+ \int_0^\infty\frac{Ai(ix)-Ai(- ix)}{e^{2\pi x}-1}\,dx}$$
$$\mathrm{A_1=\sum_0^\infty Ai(x)= \mathop=^{Abel}_{Plana}\frac12 Ai(0)+\int_0^\infty Ai(x) \,dx+\int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx= \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac13+ \int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx}$$
This means that the conjectured answer is:
$$\mathrm{A=\sum_{x\in\Bbb Z}Ai(x)=A_0+A_1-Ai(0) \mathop=^{Abel}_{Plana} \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac23+ \int_0^\infty\frac{Ai(ix)-Ai(- ix)}{e^{2\pi x}-1} \, dx + \frac{1}{3^\frac23 2Γ\left(\frac23\right)}+\frac13+ \int_0^\infty\frac{Ai(-ix)-Ai( ix)}{e^{2\pi x}-1}\,dx-Ai(0)= \frac{1}{3^\frac23 Γ\left(\frac23\right)}+ 1+\int_0^\infty\frac{Ai(ix)-Ai( -ix)+Ai(-ix)-Ai(-ix)}{e^{2\pi x}-1}\,dx-\frac{1}{3^\frac23 Γ\left(\frac23\right)}}=1 $$
My final conjecture is the following with alternate forms. Note that some simplifications are possible, but change the definition. The following also uses Hypergeometric functions:
$$\mathrm{1\mathop=^?A=\sum_{-\infty}^\infty Ai(x)=\sum_{x\in\Bbb Z}Ai(x)= \sum_{-\infty}^\infty \left(\frac{\,_0F_1\left(\frac23,\frac{x^3}{9}\right)}{3^\frac23Γ\left(\frac23\right)}-\frac{\,_0F_1\left(\frac43,\frac{x^3}{9}\right)}{\sqrt[3]3 Γ\left(\frac13\right)}\right)} =\sum_{-\infty}^\infty\left(\left(x^\frac32\right)^\frac13I_{-\frac13}\left(\frac{2x^\frac32}{3}\right)-x \left(x^\frac32\right)^{-\frac13} I_\frac13\left(\frac{2x^\frac32}{3}\right)\right)  $$
There are also variations with the Airy Bi function, and the Scorer functions Gi and Hi. Please correct me and give me feedback!
 A: I got something in terms of a theta function:
\begin{align*}
\sum\limits_{n =  - \infty }^\infty  \operatorname{Ai}(n) & =
\sum\limits_{n =  - \infty }^\infty  {\frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\left( {\tfrac{1}{3}t^3  + nt} \right)} \right)dt} } \\ &  = \frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\tfrac{1}{3}t^3 } \right)\sum\limits_{n =  - \infty }^\infty  {\exp \left( {i n   t} \right)} dt} 
\\ &
 = \frac{1}{\pi }\Re \int_0^{ + \infty } {\exp \left( {i\tfrac{1}{3}t^3 } \right)\theta _3 \!\left( {\tfrac{t}{2},1} \right)dt} \\ & = \frac{1}{\pi }\int_0^{ + \infty } {\cos \left( {\tfrac{1}{3}t^3 } \right)\theta _3 \!\left( {\tfrac{t}{2},1} \right)dt} .
\end{align*}
A: This will be a section for the “variations” mentioned in the question:
Corollary for Abel-Plana formula:
Let $f(x)$ fit the conditions for the Abel-Plana, $f(ix)-f(-ix)+f(-ix)-f(ix)=0$, and $f(0)$  exist then:
$$\sum_{\Bbb Z} f(x)=-f(0)+\sum_0^{\infty} f(-x)+\sum_0^\infty f(x)=-f(0)+\frac12 f(-0)+\int_0^\infty f(-x)dx+i\int_0^\infty \frac{f(- - ix)-f(-ix)}{e^{2\pi x}-1}dx+\frac12 f(0)+\int_0^\infty f(x) dx+i\int\frac{f(-ix)-f(ix)}{e^{2\pi x}-1}dx=-f(0)+f(0)+\int_0^\infty f(-x)+f(x) dx+i \int_0^\infty \frac{f(ix)-f(-ix)+f(-ix)-f(ix)}{e^{2\pi x}-1}dx=\int_0^\infty f(x)+f(-x) dx= \int_{\Bbb R} f(x)dx$$
Therefore a nice, not so useless, final result is:
$$\sum_{\Bbb Z} f(x)=\int\limits_{\Bbb R} f(x) dx$$
Ai variation:
Using @Gary’s result and the numerical integration shows that $A=1$. Now let us define the uregularized main integral definitions of the Airy Ai function. Note that I use $\Bbb R=[-\infty,\infty]$ and the derived corollary:
$$\sum_{x\in\Bbb Z} \int_0^\infty \cos\left(\frac{t^3}{3}+tx\right)dt= \int_{-\infty}^\infty \int_0^\infty \cos\left(\frac{t^3}{3}+tx\right)dt\,dx=\pi, \sum_{x\in\Bbb Z} \int_{\Bbb R} e^{i\left(\frac{t^3}{3}+tx\right)}dt= \iint_\limits{\Bbb R^2} e^{i\left(\frac{t^3}{3}+tx\right)}dt\,dx=2\pi$$
Bi version using @Gary’s suggestion of DLMF 9.5.3 and my finding of the Scorer function integral representations
$$\pi\sum_{-\infty}^0\text{Bi}(x)= \sum_{x=0}^\infty  \int_0^\infty e^{-\frac{t^3}{3}-tx}dt+ \sum_{x=0}^\infty \int_0^\infty \sin\left(\frac{t^3}{3}+tx\right)dt= \pi \sum_{x=0}^\infty (\text{Gi}(x)+\text{Hi}(-x))$$
