= Solution
Let $d=\operatorname{dist}(\Omega',\partial\Omega)>0$. Choose a <smooth cutoff function> $\eta\in C_c^\infty(\Omega)$ with $0\leq\eta\leq1$, $\eta=1$ on $\Omega'$ and $|D\eta|\leq C(n)/d$. Such cutoffs can be constructed by mollifying the indicator of a fixed small neighbourhood of $\overline{\Omega'}$. Write $M=\|A\|_{\infty,\mathrm{op}}$, $B=\|b\|_\infty$ and $C_0=\|c\|_\infty$; these are controlled by the individual coefficient bounds and dimension.
The <weak formulation>, extended to <compactly supported> $H^1$ tests by density, is
$$
\int_\Omega A Du\cdot D\varphi-\int_\Omega(b\cdot Du+cu)\varphi=-\int_\Omega f\varphi.
$$
We may use $\varphi=\eta^2u$, since $u\in H^1_{\mathrm{loc}}$. Expanding its <gradient> and using <uniform ellipticity> gives
$$
\lambda\int\eta^2|Du|^2\leq2M\int\eta|u||Du||D\eta|+B\int\eta^2|u||Du|+C_0\int\eta^2u^2+\int\eta^2|fu|.
$$
Apply <Young inequality> to the first two terms, allocating at most $\lambda/4$ of the <gradient> integral to each. Bound $|fu|\leq(f^2+u^2)/2$ in the last term. Absorbing those <gradient> contributions yields the <Caccioppoli inequality with bounded lower-order terms>
$$
\int\eta^2|Du|^2\leq C(n,\lambda,M,B,C_0)\left[\int(1+|D\eta|^2)u^2+\int\eta^2f^2\right].
$$
Consequently $\|Du\|_{L^2(\Omega')}\leq C(\|u\|_{L^2(\Omega)}+\|f\|_{L^2(\Omega)})$. Adding the $L^2$ <norm> of $u$ proves
$$
\boxed{\|u\|_{W^{1,2}(\Omega')}\leq C\left(\|u\|_{L^2(\Omega)}+\|f\|_{L^2(\Omega)}\right),}
$$
where $C$ has precisely the stated dependence on dimension, ellipticity, coefficient bounds and $d$. The proof uses $|c|$, so no sign condition on the zeroth-order term is required.
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