Yes, it is perfectly possible to have an embedded eigenvalue at the bottom of the spectrum. I do not have a reference (although I am quite sure there is one), but here is a simple example in dimension $d = 1$. Extension to higher dimensions is immediate.

Let
$$ \phi(x) = \frac{1}{1 + x^2} $$
and
$$ V(x) = \frac{\Delta \phi(x)}{\phi(x)} = \frac{6 x^2 - 2}{(1 + x^2)^2} \, .$$
Then:

$\phi \in L^2$, $-\Delta \phi + V \phi = 0$ (by definition of $V$), and so $\phi$ is an eigenfunction with eigenvalue $0$;

$V(x) \to 0$ as $|x| \to \pm \infty$, so that the essential spectrum of $-\Delta + V$ is $[0, \infty)$;

$-\Delta + V(x)$ has no negative eigenvalues: if there were any, then the ground state would be orthogonal to $\phi$, and so it would necessarily change sign, a contradiction with the Courant–Hilbert nodal domain theorem.

Thus, $0$ is the bottom of the spectrum of $-\Delta + V$, and it is both an eigenvalue and a point in the essential spectrum, as desired.

positiveeigenvalues only. $\endgroup$4more comments