= Solution
Put $q(x)=\sigma(x)^2p(x)$. The explicit density formula and local integrability assumption give
$$
q(x)=C\exp\left(\int_0^x\frac{2b(s)}{\sigma(s)^2}\,ds\right),\qquad q'(x)=2b(x)p(x).
$$
Hence its <Fokker-Planck probability current> $bp-q'/2$ is zero. The stationary adjoint equation is $L^*p=-(bp)'+q''/2=0$. We prove the required integral identity with cutoffs, avoiding an unstated boundary condition on derivatives of $u$.
Choose smooth $0\leq\chi_R\leq1$, equal to one on $[-R,R]$, supported in $[-2R,2R]$, with $|\chi_R'|\leq C_1/R$ and $|\chi_R''|\leq C_2/R^2$. Since $pLu=(qu_x)'/2$, integrate the backward equation in time and integrate by parts twice in space:
$$
\int\chi_Rp\,[u(t,\cdot)-f],dx
=\frac12\int_0^t\int u(s,x)[\chi_R''(x)q(x)+\chi_R'(x)q'(x)]\,dx\,ds.
$$
These integrations have compact support, as permitted in the question. Bounded coefficients and normalized $p$ give
$$
\|q\|_1\leq\|\sigma\|_\infty^2,\qquad\|q'\|_1\leq2\|b\|_\infty.
$$
The right side is bounded in absolute value by
$$
\frac{t\|u\|_\infty}{2}\left(\frac{C_2\|q\|_1}{R^2}+\frac{C_1\|q'\|_1}{R}\right),
$$
which tends to zero. <Dominated convergence> on the left therefore proves
$$
\boxed{\int_{\mathbb R}u(t,x)p(x)\,dx=\int_{\mathbb R}f(x)p(x)\,dx.}
$$
This is the <cutoff proof of invariance for a zero-flux diffusion density>. In particular, if an independent initial state has density $p$, part (a) and Fubini give a constant expectation of this test function at every time. This consequence alone is sufficient for the final coupling argument.
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