In "simple" one-dimensional cases like this, one can often solve the first-order Schrödinger equation analytically:
$$
\Big[H^{(0)}-E^{(0)}\Big]|\psi^{(1)}\rangle=-\Big[W-E^{(1)}\Big]|\psi^{(0)}\rangle
$$
as an inhomogeneous differential equation to find the exact first-order wave function. Then the second- and third-order energies can be readily computed, and the pain of dealing with scattering states is avoided, because not a single sum-over-states appears in our perturbation formulae.
To keep things simple, I start with natural units $\hbar=c=1$, so that
$$
H=-\frac{1}{2m}\frac{\mathrm{d}^2}{\mathrm{d} x^2}-\alpha\delta(x)+a\sin(kx) \ ,
$$
where $\delta(x)$ has dimension $1/\text{length}=\text{energy}$, so $\alpha$ is dimensionless. Let us introduce the energy and length scales
$$
{\cal{E}}_0=m\alpha^2 \ \ \ , \ \ \ l_0=\frac{1}{m\alpha} \ .
$$
Then the dimensionless problem reads
$$
{\cal{H}}=\frac{1}{{\cal{E}}_0}H=-\frac{1}{2}\frac{\mathrm{d}^2}{\mathrm{d} {x'}^2}-\delta(x')+q\sin(\kappa x') \ ,
$$
where $q=a/{\cal{E}}_0$, $\kappa=kl_0$ and $x'=x/l_0$ (but I will name it as $x$ from now on for simplicity). The zeroth-order solution is just
$$
\psi^{(0)}(x)=\exp(-|x|) \ \ , \ \ E^{{(0)}}=-\frac{1}{2} \ ,
$$
and $E^{{(1)}}=0$. The first-order Schrödinger equation reads
$$
\left[-\frac{1}{2}\frac{\mathrm{d}^2}{\mathrm{d} x^2}-\delta(x)+\frac{1}{2}\right]\psi^{(1)}(x)=-q\sin(\kappa x)\exp(-|x|) \ .
$$
Since $\psi^{(0)}(x)$ is even and the perturbation is odd, we expect from symmetry considerations that $\psi^{(1)}(x)$ is odd. This means $\psi^{(1)}(0)=0$, so the delta function term can be dropped:
$$
\left[-\frac{1}{2}\frac{\mathrm{d}^2}{\mathrm{d} x^2}+\frac{1}{2}\right]\psi^{(1)}(x)=-q\sin(\kappa x)\exp(-|x|) \ .
$$
The well-behaving solution for $x>0$ is
$$
\psi^{(1)}_+(x)=c_+\exp(-x)+\frac{2q}{\kappa(4+\kappa^2)}\Big[2\cos(\kappa x)-\kappa\sin(\kappa x)\Big]\exp(-x) \ ,
$$
and for $x<0$, it is
$$
\psi^{(1)}_-(x)=c_-\exp(+x)-\frac{2q}{\kappa(4+\kappa^2)}\Big[2\cos(\kappa x)+\kappa\sin(\kappa x)\Big]\exp(+x) \ .
$$
Matching the two solutions as $x\rightarrow0$ leads to the relation
$$
c_+-c_-=-\frac{8q}{\kappa(4+\kappa^2)} \ .
$$
We need one more condition to completely fix the first-order wave function, the intermediate normalization:
$$
\langle\psi^{(0)}|\psi^{(1)}\rangle=\int_{-\infty}^0\mathrm{d}x\psi^{(0)}(x)\psi^{(1)}_-(x)+\int_0^{+\infty}\mathrm{d}x\psi^{(0)}(x)\psi^{(1)}_+(x)=0 \ .
$$
After carrying out the integrals, we find $c_++c_-=0$ from this condition, so
$$
c_\pm=\mp\frac{4q}{\kappa(4+\kappa^2)} \ ,
$$
and $\psi^{(1)}(x)$ is completely determined:
$$
\psi^{(1)}(x)=-\frac{2q}{4+\kappa^2}\text{sgn}(x)\left[2\frac{1-\cos(\kappa x)}{\kappa}+\sin(\kappa |x|)\right]\exp(-|x|) \ .
$$
The solution has odd parity and vanishes at $x=0$, as expected.
The second-order energy is then
$$
E^{(2)}=\langle\psi^{(0)}|W|\psi^{(1)}\rangle=-\frac{\kappa^2(10+\kappa^2)}{(1+\kappa^2)(4+\kappa^2)^2}q^2 \ ,
$$
manifestly negative, as all second-order corrections to ground state energies should be. Because of the $2n+1$ theorem of Wigner, knowing $\psi^{(1)}(x)$ would be sufficient to determine the third-order energy as well:
$$
E^{(3)}=\langle\psi^{(1)}|W|\psi^{(1)}\rangle-E^{(1)}\langle\psi^{(1)}|\psi^{(1)}\rangle \ ,
$$
but this is of course zero due to parity reasons, just like $E^{(1)}$ was.
So in conclusion, the energy of the perturbed system seems to be
$$
E=-\frac{1}{2}-\frac{\kappa^2(10+\kappa^2)}{(1+\kappa^2)(4+\kappa^2)^2}q^2+{\cal{O}}(q^4) \ .
$$
This system is very special because the unperturbed Hamiltonian has only a single bound state, so all perturbation contributions come from the continuum part of the zeroth-order spectrum. But actually, continuum states are always important in perturbation theory, even if one has an infinite number of zeroth-order bound states (like in the case of the hydrogen atom). I prepared some examples of this (probably surprising) phenomenon in an earlier answer.
