Let $A$ be a self-adjoint operator with domain $D(A)\subset\mathcal H$ ($\mathcal H$ is some Hilbert space). An operator $C$ with $D(A)\subset D(C)$ is called relatively compact with respect to $C$ if $C(A-zI)^{-1}$ is compact for some (hence all) $z\notin\sigma(A)$. Paraphrasing Corollary 2, page 113 Section XIII.4, in [1], we have

If $C$ is relatively compact with respect to $A$, then $B:=A+C$ is closed with domain $D(B)=D(A)$, and $$\sigma_{ess}(B)=\sigma_{ess}(A).$$

In fact, this is elementary once one realises that $C(A-zI)^{-1}:\mathcal H\to \mathcal H$ is compact if and only if $C:D(A)\to\mathcal H$ is compact.

In your case, setting $A = -\partial_x^2 + c_*$ and $C$ the multiplication operator by $\Phi$, is it not hard to prove that $C:H^2(\mathbb R)\to L^2(\mathbb R)$ is compact. We then obtain that
$$
\sigma_{\mathrm{ess}}(\mathcal{L}) = \sigma_{\mathrm{ess}}(-\partial_x^2 + c_*).
$$
Since it is well know that $\sigma_{\mathrm{ess}}(-\partial_x^2 + c_*) = [c_*,+\infty)$, the result follows.

For your second question, this answer is yes the result hold for $\Phi(x) = e^{-|x|}$. In fact it will hold in dimension $d\leq 3$ for any $\Phi\in L^2(\mathbb R^d)$. As explained earlier, it is enough to show that multiplication by $\Phi$ is (well-defined and) compact from $H^2(\mathbb R^d)\to L^2(\mathbb R^d)$; let us prove this result.

Let $(f_n)_{n\geq0}$ be a bounded sequence in $H^2(\mathbb R^d)$, $d\leq 3$. The core of the argument is the following fact.

$(f_n)_{n\geq0}$ is bounded in $L^\infty$ and, up to extraction, converges $L^\infty$-locally to some function $f$.

Note that $f$ then has to be bounded as well. Once this is established, we see that
$$ \begin{align*}
\limsup |\Phi f_n-\Phi f|_{L^2}^2
& = \limsup \int|\Phi|^2\cdot|f_n-f|^2 \\\\
& \leq \limsup |f_n-f|_{L^\infty([-k,k]^d)}^2\cdot\int|\Phi|^2{\mathbf 1}_{[-k,k]^d} \\\\
& \quad + \limsup (|f_n|_{L^\infty}+|f|_{L^\infty})^2\cdot\int|\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement} \\\\
& \leq |\Phi|_{L^2}^2\cdot0
+ 4M^2\cdot\limsup \int |\Phi|^2{\mathbf 1}_{([-k,k]^d)^\complement}
\end{align*} $$
for $M$ a bound on the norms $|f_n|_{L^\infty}$, and $\limsup$ the limit superior along a convergent subsequence. Because $\Phi$ is in $L^2$, Lebesgue's dominated convergence theorem guarantees that the limit is zero, and $\Phi f_n$ converges to $\Phi f$ as expected.

Let us turn to the proof of the fact. The boundedness of $(f_n)_{n\geq0}$ on compact sets follows from the continuous embeddings from $H^2(\ell+[0,1]^d)$ to $\mathcal C^0(\ell+[0,1]^d)$ for all $\ell\in\mathbb Z^d$ (because $d\leq3$). Since the norm of these injections does not depend on $\ell$, $(f_n)_{n\geq0}$ is in fact uniformly bounded over $\mathbb R^d$

As for the convergence up to extraction, according to the usual Sobolev embeddings/inequalities, the sequence of restrictions $\big((f_n)_{|[-k,k]^d}\big)_{n\geq0}$ is relatively compact in $\mathcal C^0([-k,k]^d)$ for all $k\geq1$ (we use again that $d\leq3$). We can then extract diagonally a subsequence $(f_{\sigma(m)})_{m\geq0}$ that converges to some continuous function $f$ uniformly over the compact sets, concluding the proof of the fact.

[1] Reed, M., Simon, B. (1978). *Methods of Modern Mathematical Physics. IV Analysis of Operators*. New York: Academic Press.