= Solution
For the odd channel, solve on $x>0$ with $\psi(0)=0$. Inside $0<x<a$, write $\psi=C\sin(kx)$; outside write $\psi=e^{-ikx}-S_{--}e^{ikx}$, respecting the outgoing odd-channel minus sign. <Continuity> at $a$ and the delta-potential <derivative> jump are
$$
C\sin(ka)=e^{-ika}-S_{--}e^{ika},\qquad
-ik(e^{-ika}+S_{--}e^{ika})-Ck\cos(ka)=U_0(e^{-ika}-S_{--}e^{ika}).
$$
Eliminating $C$ yields
$$
\boxed{S_{--}(k)=e^{-2ika}\frac{(2k-iU_0)e^{ika}+iU_0e^{-ika}}{(2k+iU_0)e^{-ika}-iU_0e^{ika}}.}
$$
The equations remain valid when $\sin(ka)=0$ by taking the limiting value, rather than dividing blindly. For real $k,U_0$, numerator and denominator inside the fraction are complex conjugates and the prefactor has unit modulus. Thus $|S_{--}|^2=1$. This is conservation of flux in the odd <parity> channel: the real potential causes only a phase shift there, not absorption. It does not assert unit left-to-right transmission in the usual incident-from-one-side basis.
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