= Solution
Away from $x=\xi$, a <Green function> for this second-derivative operator is linear in $x$. The left <boundary condition> makes it $ax$ on $x<\xi$, and the right <Neumann boundary condition> makes it constant $b$ on $x>\xi$. Continuity gives $b=a\xi$. Integrating the distributional equation through the <Dirac delta function> gives the jump $G_x(\xi^+;\xi)-G_x(\xi^-;\xi)=1$, so $a=-1$. Therefore
$$
\boxed{G(x;\xi)=-\min(x,\xi)=
\begin{cases}-x,&x\leq\xi,\\-\xi,&x\geq\xi.\end{cases}}
$$
Apply the <Green function> to the forcing:
$$
y(x)=\int_0^1G(x;\xi)\xi e^{-\xi}\,d\xi
=-\int_0^x\xi^2e^{-\xi}\,d\xi-x\int_x^1\xi e^{-\xi}\,d\xi.
$$
Evaluation yields
$$
\boxed{y(x)=(x+2)e^{-x}+\frac{2x}{e}-2.}
$$
Indeed $y'(x)=-(x+1)e^{-x}+2/e$, so $y''=xe^{-x}$, $y(0)=0$ and $y'(1)=0$. A difference of two solutions is linear with these homogeneous boundary conditions, hence zero; the <boundary value problem> has a unique solution.
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