To maintain the meaning of this question, $a\neq0$ and $s\neq0$ should be restricted
Let $f(x)=\int_Ce^{xu}K(u)~du$ ,
Then $(\int_Ce^{xu}K(u)~du)''+2ax(\int_Ce^{xu}K(u)~du)'-2s\int_Ce^{xu}K(u)~du=0$
$\int_Cu^2e^{xu}K(u)~du+2ax\int_Cue^{xu}K(u)~du-2s\int_Ce^{xu}K(u)~du=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2aue^{xu}K(u)~d(xu)=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+\int_C2auK(u)~d(e^{xu})=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}~d(2auK(u))=0$
$\int_C(u^2-2s)e^{xu}K(u)~du+[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)+2aK(u))~du=0$
$[2aue^{xu}K(u)]_C-\int_Ce^{xu}(2auK'(u)-(u^2-2a-2s)K(u))~du=0$
$\therefore 2auK'(u)-(u^2-2a-2s)K(u)=0$
$2auK'(u)=(u^2-2a-2s)K(u)$
$\dfrac{K'(u)}{K(u)}=\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}$
$\int\dfrac{K'(u)}{K(u)}du=\int\left(\dfrac{u}{2a}+\left(-\dfrac{s}{a}-1\right)\dfrac{1}{u}\right)du$
$\ln K(u)=\dfrac{u^2}{4a}+\left(-\dfrac{s}{a}-1\right)\ln u+c_1$
$K(u)=cu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}}$
$\therefore f(x)=\int_Ccu^{-\frac{s}{a}-1}e^{\frac{u^2}{4a}+xu}~du$
But since the above procedure in fact suitable for any complex number $u$ ,
$\therefore f_n(x)=\int_{a_n}^{b_n}c_n(m_nt)^{-\frac{s}{a}-1}e^{\frac{(m_nt)^2}{4a}+xm_nt}~d(m_nt)={m_n}^{-\frac{s}{a}}c_n\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$
For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:
$\lim\limits_{t\to a_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}=\lim\limits_{t\to b_n}t^{-\frac{s}{a}}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}$
$\int_{a_n}^{b_n}t^{-\frac{s}{a}-1}e^{\frac{{m_n}^2t^2}{4a}+m_nxt}~dt$ converges
Case $1$ : $a>0$ and $s<0$
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm i$
$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt$
Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\sin xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{-\frac{t^2}{4a}}\cos xt~dt$
Case $2$ : $a<0$ and $s>0$
For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm1$
$\therefore f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$ or $C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt$
Hence $f(x)=C_1\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\sinh xt~dt+C_2\int_0^\infty t^{-\frac{s}{a}-1}e^{\frac{t^2}{4a}}\cosh xt~dt$
Compare with $\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ and $\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)$ , according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_functions#Integral_representations,
$\Phi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^1t^{-\frac{s}{2a}-1}(1-t)^{\frac{s}{2a}-\frac{1}{2}}e^{-ax^2t}~dt$ when $\dfrac{1}{2}>-\dfrac{s}{2a}>0$
$\Psi\left(-\dfrac{s}{2a},\dfrac{1}{2},-ax^2\right)\propto\int_0^\infty t^{-\frac{s}{2a}-1}(1+t)^{\frac{s}{2a}-\frac{1}{2}}e^{ax^2t}~dt$ when $-\dfrac{s}{2a}>0$