You've got two questions about the absolute Laplace transform, which is defined as
$$\mathcal L_a[f(k)](x)=\int^\infty_{0}f(k)e^{-k|x|}dk$$ for continuous $f(k) \in o(e^{\delta x}), \forall\delta>0$.
Question 1: How to interpret the different behaviour on both sides of the equation under differentiation at $x=0$?
In $\mathbb R$, if there exists a punctured neighbourhood of $x=0$ such that $\mathcal L_a[f](x)$ and $\displaystyle{\sum^\infty_{n=0}a_n x^{2n}}$ coincide , then both expressions have the same behaviour under differentiation at $x=0$, in the sense that
$$\int^\infty_{0}f(k) \left( \frac{\partial}{\partial x} \right)_{\text{left}} e^{-k|x|}dk\bigg\vert_{x\to0^-}
=\int^\infty_{0}f(k) \left( \frac{\partial}{\partial x} \right)_{\text{right}} e^{-k|x|}dk\bigg\vert_{x\to0^+}
=\frac{d}{dx}\sum^\infty_{n=0}a_n x^{2n}\bigg\vert_{x=0}=0
$$
where the subscripts denote the one-sided derivatives.
Sketch of proof:
The third equality is trivial.
To prove the right-sided derivative is zero, we want to switch the order of integral and differentiation. Here we will utilise the 'extended Leibniz integral rule':
For $$\frac{d}{dx}\int^\infty_c f(x,t)dt=\int^\infty_c \frac{\partial}{\partial x} f(x,t)dt\qquad{x\in(a,b)}$$ to be true, the sufficient conditions are
$f(x,t)$ and $\displaystyle{\frac{\partial}{\partial x} f(x,t)}$ are continuous in the region $c\le t<\infty$, $a\le x\le b$.
$\displaystyle{\lim_{N\to\infty}\int^N_c \frac{\partial}{\partial x}f(x,t)dt} $ converges uniformly for $x\in(a,b)$.
$\displaystyle{\lim_{N\to\infty}\int^N_c f(x,t)dt} $ converges for $x\in(a,b)$.
It is straight-forward to prove that the three conditions are satisfied for $0<x<r$ ($r$ is the radius of convergence of the Taylor series). Thus
$$\begin{align}
\int^\infty_{0}f(k) \left( \frac{\partial}{\partial x} \right)_{\text{right}} e^{-k|x|}dk\bigg\vert_{x\to0^+}
&=\left( \frac{d}{dx} \right)_{\text{right}}\int^\infty_{0}f(k)e^{-kx}dk\bigg\vert_{x\to0^+} \\
&=\left( \frac{d}{dx} \right)_{\text{right}}\sum^\infty_{n=0}a_n x^{2n}\bigg\vert_{x\to0^+} \\
&=0
\end{align}
$$
Similarly, the left-sided derivative is also zero.
Note: It is a little bit more complicated to show condition 2 is satisfied.
We aim to prove that for $x>0$, $$\lim_{N\to\infty}\int^N_{0}f(k) \left( \frac{\partial}{\partial x} \right)_{\text{right}} e^{-k|x|}dk=\lim_{N\to \infty}\int^N_{0}-kf(k) e^{-kx}dk\quad\text{converges uniformly.}$$
To this end, we make use of Cauchy criterion:
for sufficiently large $m>n>N$,
$$\begin{align}
\left|\int^m_{n}-kf(k) e^{-kx}dk\right|
&<\int^m_{n}\left|kf(k) e^{-kx}\right|dk \\
&<\int^m_{n}e^{\delta x} e^{-kx}dk \\
&<2\cdot\frac{e^{(\delta-x)n}}{x-\delta} \\
&<2\cdot\frac{e^{(\delta-x)N}}{x-\delta} \\
&<2\cdot\frac{e^{-\Delta N}}{\Delta} \quad \text{for } x>\delta+\Delta, \Delta>0\\
\end{align}
$$
Choosing $N=\frac1{\Delta}\ln\frac 2{\epsilon\Delta}$ would show uniform convergence for $x>\delta+\Delta$, and hence justifying the exchange of differentiation and integral for $x>\delta+\Delta$. By noticing that $\delta,\Delta$ can be chosen to be arbitrarily small, we have shown that condition 2 is satisfied for all $x>0$.
Question 2: Is it possible to differentiate under the integral sign, possibly at the cost of introducing additional terms of distributional nature?
Yes.
Suppose $\mathcal L_a[f](x)$ and $\displaystyle{\sum^\infty_{n=0}a_n x^{2n}}$ coincide in a punctured neighbourhood of $x=0$.
Then, indeed, in the sense of distribution,
$$\int^\infty_0 kf(k)dk=0$$ and thus
$$\frac{d}{dx}\mathcal L_a[f(k)](x)\bigg\vert_{x=0}=-\text{sgn}(x)\int^\infty_0 kf(k)e^{-k|x|}dk\bigg\vert_{x=0}=-\text{sgn}(0)\int^\infty_0 kf(k)dk=0$$
Proof:
It is well-known that $$\int^\infty_0 \delta'(s)e^{-sk}ds=k$$
Therefore,
$$\begin{align}
\int^\infty_0 kf(k)dk
&=\int^\infty_0 \int^\infty_0 \delta'(s)e^{-sk} f(k) \, ds \, dk \\
&=\int^\infty_0 \int^\infty_0 \delta'(s)e^{-sk} f(k) \, dk \, ds \qquad (1)\\
&=\int^\infty_0 \delta'(s)\left(\int^\infty_0 f(k)e^{-sk} dk\right)ds \\
&=\int^\infty_0 \delta'(s)\sum^\infty_{n=0}a_n s^{2n} ds \qquad (2)\\
&=-\int^\infty_0 \delta(s)\left(\sum^\infty_{n=0}a_n s^{2n}\right)' ds \\
&=-\left(\sum^\infty_{n=0}a_n s^{2n}\right)'_{s=0} \\
&=0
\end{align}
$$
$(1)$: Changing order of integrals is justified by Fubini's theorem.
$(2)$: Due to the formula $\displaystyle{\int^\infty_{-\infty}\delta'(x)\varphi(x)dx=-\int^\infty_{-\infty}\delta(x)\varphi'(x)dx}$.
Introduce the function
$$
f(x,t)=I_0(x+t)K_1(x)-K_0(x+t)I_1(x).
$$
We want to show that $f(x,t)>0$ for all $x>0,t\geq 0$.
Consider first the case $f(x,0)$. We can prove that $f(x,0)>0$ by writing
$$
f(x,0)=(I_0(x)-I_1(x))K_1(x)+I_1(x)(K_1(x)-K_0(x))
$$
and using the monotonicity properties (see here: https://dlmf.nist.gov/10.37)
$$
I_0(x)>I_1(x)\mbox{ and }K_1(x)>K_0(x)\mbox{ for all }x>0.
$$
Next, by a direct computation, we have
$$
\frac{\partial}{\partial t}f(x,t) =I_1( t + x) K_1(x) + I_1(x)K_1(t + x)
$$
which is manifestly positive for $x>0,t\geq 0$. Hence
$$
f(x,t)=f(x,0)+\int_0^t (I_1( \tau+ x) K_1(x) + I_1(x)K_1(\tau + x))\mathrm d \tau>0.
$$
Best Answer
We may use the Hankel transform representation of the Gaussian term (DLMF): \begin{equation} \int_{0}^{\infty}J_{\nu}\left(bt\right)\exp\left(-p^{2}t^{2}\right)t^{\nu+1}% \mathrm{d}t=\frac{b^{\nu}}{(2p^{2})^{\nu+1}}\exp\left(-\frac{b^{2}}{4p^{2}}% \right) \end{equation} with $\nu=0,p=1/2$ to express (by changing the order of integration) \begin{align} I(x)&=\int_0^\infty k e^{-k^2} J_0(kx)Y_0(kx)~\mathrm{d}k \\ &=\frac{1}{2}\int_0^\infty te^{-t^2/4}\,dt\int_0^\infty k J_0(kt) J_0(kx)Y_0(kx)~\mathrm{d}k \end{align} The integral of the triple product of Bessel function is (for a proof see Watson, "Theory of Bessel functions", ยง13.46): \begin{equation} \int_{0}^{\infty}Y_{\nu}\left(ak\right)J_{\nu}\left(bk\right)J_{\nu}\left(ck% \right)k^{1+\nu}\mathrm{d}k=\begin{cases}-\dfrac{(abc)^{\nu}(-A)^{-\nu-\frac{1% }{2}}}{\pi^{\frac{1}{2}}2^{\nu+1}\Gamma\left(\frac{1}{2}-\nu\right)},&0<a<|b-c% |,\\ 0,&|b-c|<a<b+c,\\ \dfrac{(abc)^{\nu}(-A)^{-\nu-\frac{1}{2}}}{\pi^{\frac{1}{2}}2^{\nu+1}\Gamma% \left(\frac{1}{2}-\nu\right)},&a>b+c.\end{cases} \end{equation} where $A=s(s-a)(s-b)(s-c), s=(a+b+c)/2$. Here, we choose $\nu=0,a=x,b=t,c=x$, and thus $A=-\left( t^2/4-x^2 \right)t^2/4$. This integral is non zero iff $t>2x$. In this case, \begin{equation} \int_{0}^{\infty}Y_{\nu}\left(ak\right)J_{\nu}\left(bk\right)J_{\nu}\left(ck% \right)k^{1+\nu}\mathrm{d}k=-\frac{t^{-1}\left( t^2/4 -x^2\right)^{-1/2}}{\pi} \end{equation} Using simple changes of variables and an integral representation of the modified Bessel function (DLMF), we have then \begin{align} I(x)&=-\frac{1}{2\pi}\int_{2x}^\infty \frac{e^{-t^2/4}}{\sqrt{t^2/4-x^2}}\,dt\\ &=\frac{-1}{\pi}\int_0^\infty e^{-x^2\cosh^2u}du\\ &=-\frac{e^{-x^2/2}}{\pi}\int_0^\infty e^{-\frac{x^2}{2}\cosh 2u}\,du\\ &=-\frac{e^{-x^2/2}}{2\pi}\int_0^\infty e^{-\frac{x^2}{2}\cosh v}\,dv\\ &=-\frac{e^{-x^2/2}}{2\pi}K_0\left( \frac{x^2}{2} \right) \end{align} as expected.