The correct mathematical way to show this is to take a test function $\varphi\in C^\infty_c(\mathbb R)$ and show that
$$
\int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx = \varphi(0),
$$
where $L^* = \frac{d^{2}}{d x^{2}}-a \frac{d}{d x}+b.$
Inserting the definition of $F$ into the left hand side gives
$$\begin{align}
\int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx
&= \int_{-\infty}^{0} g(x) \, \left(\varphi''(x)-a\varphi'(x)+b\varphi(x)\right) \, dx\\
&+ \int_{0}^{\infty} f(x) \, \left(\varphi''(x)-a\varphi'(x)+b\varphi(x)\right) \, dx
.
\end{align}$$
Integrating by parts, moving derivatives from $\varphi$ to $g$ and $f$ we get
$$\begin{align}
\int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx
&= \left[ g(x)\,\varphi'(x) - g'(x)\,\varphi(x) - g(x)\,\varphi(x) \right]_{-\infty}^{0} \\
&+ \int_{-\infty}^{0} \left( g''(x)+ag'(x)+bg(x)\right) \, \varphi(x) \, dx \\
&+ \left[ f(x)\,\varphi'(x) - f'(x)\,\varphi(x) - f(x)\,\varphi(x) \right]_{0}^{\infty} \\
&+ \int_{0}^{\infty} \left( f''(x) + af'(x) + bf(x) \right) \, \varphi(x) \, dx
.
\end{align}$$
Here the integrals vanish and the $[\ldots]$ parts vanish at $\pm\infty$ so we are left with
$$\begin{align}
\int_{-\infty}^{\infty} F(x) \, L^*\varphi(x) \, dx
&= \left( g(0) \, \varphi'(0) - g'(0) \, \varphi(0) - g(0) \, \varphi(0) \right) \\
&- \left( f(0) \, \varphi'(0) - f'(0) \, \varphi(0) - f(0) \, \varphi(0) \right) \\
&= \left( g(0)-f(0) \right) \varphi'(0) - \left( g'(0)-f'(0) \right) \varphi(0) - \left( g(0) - f(0) \right) \varphi(0) \\
&= 0 \varphi'(0) - (-1)\varphi(0) - 0\varphi(0) \\
&= \varphi(0).
\end{align}$$