= Solution
After the shift $x(\tau)=x_0(\tau)+y(\tau)$, the endpoint values of $y$ are both zero. Expanding the <Euclidean action> gives
$$
S[x_0+y]=S[x_0]+S[y]
+m\int_0^{\mathcal T}(\dot x_0\dot y+\omega^2x_0y)\,d\tau.
$$
By <integration by parts>, the cross term is
$$
m[\dot x_0y]_0^{\mathcal T}
+m\int_0^{\mathcal T}(-\ddot x_0+\omega^2x_0)y\,d\tau=0.
$$
The first term vanishes by the homogeneous <Dirichlet boundary conditions>, and the second by the <Euler-Lagrange equation>. The translation has unit <Jacobian determinant> in the <time-sliced configuration-space path integral>, so $\mathcal D x=\mathcal D y$. Consequently <classical-path factorization of a quadratic path integral> is exact here:
$$
\boxed{\langle x|e^{-\beta H}|x\rangle=e^{-S[x_0]/\hbar}Z,\qquad
Z=\int_{y(0)=y(\mathcal T)=0}\mathcal D y\,
\exp\!\left[-\frac m{2\hbar}\int_0^{\mathcal T}(\dot y^2+\omega^2y^2)\,d\tau\right].}
$$
All dependence on the common endpoint $x$ lies in
$$
\boxed{e^{-S[x_0]/\hbar}
=\exp\!\left[-\frac{m\omega x^2}{\hbar}\tanh\!\left(\frac{\beta\hbar\omega}{2}\right)\right],
\qquad Z\text{ is independent of }x.}
$$
The endpoint domain and the integrand defining $Z$ are the same for every $x$. This $Z$ is a zero-endpoint <Gaussian path integral>; the <canonical partition function> instead integrates the entire diagonal kernel over $x$. The final sentence of the printed part omits $/\hbar$ from its exponential: the physical <Boltzmann factor> has the dimensionless exponent $-S/\hbar$.
An optional normalization check uses the <Dirichlet oscillator fluctuation determinant>. Its <eigenvalues> are $m[(n\pi/\mathcal T)^2+\omega^2]$, $n\ge1$. Relative to the free-particle <Gaussian path integral>, the <hyperbolic-sine infinite product> gives
$$
\frac{Z}{\sqrt{m/(2\pi\hbar\mathcal T)}}
=\prod_{n=1}^{\infty}\left[1+\frac{\omega^2\mathcal T^2}{n^2\pi^2}\right]^{-1/2}
=\sqrt{\frac{\omega\mathcal T}{\sinh(\omega\mathcal T)}}.
$$
Thus $Z=\sqrt{m\omega/[2\pi\hbar\sinh(\beta\hbar\omega)]}$. Integrating the resulting <diagonal thermal kernel of a quantum harmonic oscillator> by a <Gaussian integral> yields $\operatorname{Tr}e^{-\beta H}=[2\sinh(\beta\hbar\omega/2)]^{-1}$, including the <zero-point energy>.
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