= Solution
In the <Parity basis of a one-dimensional S-matrix>, parity acts as $\mathcal P=\operatorname{diag}(1,-1)$. <Parity> invariance requires
$$
[S,\mathcal P]=0,
$$
and therefore
$$
S_{+-}=S_{-+}=0.
$$
Independently, <unitary operator>[unitarity] gives
$$
|S_{++}|^2+|S_{-+}|^2=1,
\qquad
|S_{--}|^2+|S_{+-}|^2=1,
\qquad
S_{++}^*S_{+-}+S_{-+}^*S_{--}=0.
$$
Combining parity and unitarity leaves two phases,
$$
\boxed{S=\operatorname{diag}(e^{2i\delta_+},e^{2i\delta_-})}.
$$
For the <symmetric double-delta potential>, consider the odd channel on $x\geq0$. Choose the asymptotic convention
$$
\psi(x)=e^{-ikx}-S_{--}e^{ikx},\qquad x>a,
$$
and write $psi(x)=C\sin(kx)$ for $0<x<a$; the sine enforces odd <parity> at the origin. Continuity and the <delta potential> derivative jump at $x=a$ give
$$
C\sin(ka)=e^{-ika}-S_{--}e^{ika},
$$
$$
-ik e^{-ika}-ikS_{--}e^{ika}-Ck\cos(ka)
=U_0C\sin(ka).
$$
Eliminating $C$ yields the <Odd-parity S-matrix of a symmetric double-delta potential>
$$
S_{--}=e^{-2ika}
\frac{k e^{ika}+U_0\sin(ka)}
{k e^{-ika}+U_0\sin(ka)}.
$$
Using $2i\sin(ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expression
$$
\boxed{
S_{--}=e^{-i2ka}
\frac{(2k-iU_0)e^{ika}+iU_0e^{-ika}}
{(2k+iU_0)e^{-ika}-iU_0e^{ika}}.
}
$$
For real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.
A <bound-state poles of a partial-wave S-matrix>[bound state] occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero gives
$$
\kappa e^{\kappa a}+U_0\sinh(\kappa a)=0,
$$
or
$$
\boxed{2\kappa=-U_0(1-e^{-2\kappa a}).}
$$
For $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly when
$$
\boxed{-U_0a>1.}
$$
This is the <odd bound state of a symmetric double-delta potential>; equality is its zero-energy threshold.
For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. Put
$$
z=ka,qquad G=U_0a.
$$
It becomes
$$
z e^{-iz}+G\sin z=0.
$$
When $G\gg1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi+\eta$ and expanding gives
$$
(\pi+\eta)e^{-i\eta}+G\eta=0,
$$
so
$$
\eta=-\frac\pi G+\frac{\pi-i\pi^2}{G^2}+O(G^{-3}).
$$
In particular, the <odd resonance of a strongly repulsive symmetric double-delta potential> has
$$
\boxed{
k_R=\frac\pi a\left(1-\frac1{U_0a}+O((U_0a)^{-2})\right),
\qquad
k_I=-\frac{\pi^2}{U_0^2a^3}+O((U_0a)^{-3}/a).
}
$$
The negative imaginary part identifies a decaying <scattering resonance>. If
$$
E=\frac{\hbar^2k^2}{2m}=E_R-\frac{i\Gamma}{2},
$$
then, to leading nonzero order,
$$
\Gamma=-2\operatorname{Im}E
=\frac{2\hbar^2\pi^3}{mU_0^2a^4}.
$$
Its time-dependent amplitude is proportional to
$$
e^{-iEt/\hbar}=e^{-iE_Rt/\hbar}e^{-\Gamma t/(2\hbar)},
$$
so its survival probability decays as $e^{-\Gamma t/\hbar}$ and its lifetime is
$$
\boxed{\tau=\frac\hbar\Gamma
=\frac{mU_0^2a^4}{2\hbar\pi^3}.}
$$
Physically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by <quantum tunnelling>.
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