What is the inverse of an integral transformation that turns second order ODEs into first order ones?

Question

Let $\mathcal{S}_f:\mathcal{C^\infty\rightarrow C^\infty}$ be an integral transform such that, for any $\{f,\psi\}\subseteq\mathcal{C}^\infty$, $\int_0^\infty f(x)dx=\infty$, we have that: $$\mathcal{S}_f\{f\cdot\psi\}(t)=\int_0^\infty f(x)\psi(x)K_f(x,t)dx$$ Where $K_f$ is the kernel defined by: $$K_f(x,t)=\exp\left(-t\cdot\int_0^xf(s)ds\right)$$ By integration by parts, one can show that the linear, second order ODE: $$\ddot{y}+f(x)y=0$$ Turns into a non-linear, first order one when transformed by $\mathcal{S}_f$: $$\mathcal{S}_f\{\ddot{y}+f(x)y\}(t)=\mathcal{S}_f\{\dot{y}(tf(x)+\frac{1}{t})\}(t)+\frac{y(0)}{t}-\dot{y}(0)$$ But this non-linear equation could be easily linearized if we find an inverse integral transform $\mathcal{S}^{-1}_f:\mathcal{C^\infty\rightarrow C^\infty}$ of $\mathcal{S}_f$, this is, an integral transform that, for any valid $\psi\in\mathcal{C}^\infty$, shows: $$\mathcal{S}_f^{-1}\{\mathcal{S}_f\{\psi\}(t)\}(x)=\psi(x)$$ This can be seen by imposing $\mathcal{S}_f\{\psi\}(t)|_{t=p}$, $p\in\mathbb{C}$, which will turn the original equation into: $$(pf(x)+\frac{1}{p})\dot{y}+\mathcal{S}^{-1}_f\{\frac{y(0)}{p}-\dot{y}(0)\}(x)=0$$ Or making the appropriate substitutions: $$\alpha(x)\dot{y}+\beta(x)=0$$ But, how can i find $\mathcal{S}^{-1}_f$? Does it exist? Any suggestion will be appreciated as an answer.