= Solution
Choose the homogeneous solutions $u(x)=x\ln x$, which vanishes at $1$, and $v(x)=x(\ln x-1)$, which vanishes at $e$. Their <Wronskian> in the order $uv'-u'v$ is $x$. For a source point $1<\xi<e$, seek the <Green's function> as
$$
G(x,\xi)=\begin{cases}A(\xi)u(x),&x<\xi,\\B(\xi)v(x),&x>\xi.\end{cases}
$$
It is continuous at $x=\xi$; otherwise its second <distributional derivative> would contain a derivative of the <Dirac delta function>, which the source does not contain. Integrating the differential equation across the source, using continuity and the leading coefficient one, gives the jump condition
$$
G_x(\xi^+,\xi)-G_x(\xi^-,\xi)=1.
$$
Continuity and this derivative jump require
$$
A u(\xi)=Bv(\xi),\qquad Bv'(\xi)-Au'(\xi)=1.
$$
Solving with the <Wronskian> $\xi$ gives $A=v(\xi)/\xi=\ln\xi-1$ and $B=u(\xi)/\xi=\ln\xi$. Therefore
$$
\boxed{G(x,\xi)=\begin{cases}x\ln x(\ln\xi-1),&1\leq x\leq\xi,\\x(\ln x-1)\ln\xi,&\xi\leq x\leq e.\end{cases}}
$$
The two values at $x=\xi$ are identical. The derivative jump is explicitly
$$
(\ln\xi)\ln\xi-(\ln\xi+1)(\ln\xi-1)=1.
$$
Both endpoint values vanish and the differential expression is zero off the diagonal, completing the verification of the <Green's function>.
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