= Solution
Expand $G(x,x')$ in a complete <orthonormal> <eigenbasis> in its first argument. Applying the operator yields $\sum_n(E_n-E)c_n\varphi_n(x)=\delta(x-x')$. Completeness expands the <Dirac delta distribution> as $\sum_n\varphi_n(x)\varphi_n^*(x')$, so
$$
\boxed{c_n=\frac{\varphi_n^*(x')}{E_n-E},\qquad
G(x,x')=\sum_n\frac{\varphi_n(x)\varphi_n^*(x')}{E_n-E}.}
$$
Consequently $\lim_{E\to E_n}(E_n-E)G=\varphi_n(x)\varphi_n^*(x')$.
For the explicit formula, use $E_n-E\simeq-(\hbar^2k_n/m)(k-k_n)$ and $\sin ka\simeq(-1)^na(k-k_n)$. Also
$\sin[k_n(a/2-x)]=(-1)^{n+1}\sin[k_n(x+a/2)]$. The resulting limit is
$$
\boxed{\varphi_n(x)\varphi_n^*(x')=
\frac2a\sin\frac{n\pi(x+a/2)}a\sin\frac{n\pi(x'+a/2)}a.}
$$
The expression is symmetric in $x,x'$ and holds on either side of the source. The usual complex-analysis <residue> with respect to $E-E_n$ has the opposite sign, since the source requests the limit with $E_n-E$.
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