TL;DR: The supersymmetric partner potential to OP's potential is the constant potential, which is clearly reflectionless.
Define for later convenience the constant $\kappa:=\hbar/\sqrt{2m}$. The constant potential and OP's potential are just the two first cases ($\ell=0$ and $\ell=1$) in an infinite sequence of reflectionless attractive$^1$ potentials
$$ V_{\ell}(x)~:=~-\frac{(\kappa a)^2 \ell(\ell+1)}{\cosh^2 ax}, \qquad \ell~\in~\mathbb{N}_{0}. \tag{1}$$
Let us next consider a sequence of two superpartner potentials
$$ V_{\pm,\ell}(x)~:=~ (\kappa a)^2\left( \ell^2 -\frac{\ell(\ell \mp 1)}{\cosh^2 ax}\right). \tag{2}$$
Note that reflection properties are not altered by simultaneously shifting the potentials $V_{\ldots}$ and the energy-level $E$ upwards or downwards with an overall constant. That's just a matter of choosing a zero-point level. So, from now on, we will identify two potentials iff they differ by an overall constant. E.g., the three potentials
$$ V_{+,\ell +1} ~\sim~ V_{-,\ell}~\sim~V_{\ell} \tag{3}$$
only differ by overall constants.
The two superpartner potentials $(2)$ satisfy
$$ V_{\pm,\ell} ~=~ W_{\ell}^2 \pm \kappa W_{\ell}^{\prime} , \tag{4}$$
where
$$ W_{\ell}(x)~:=~\ell \kappa a \tanh ax \tag{5} $$
is the superpotential. One may show under fairly broad assumptions$^2$ that two superpartner TISEs$^3$
$$ -\kappa^2 \psi^{\prime\prime} +V_{\pm,\ell}\psi~=~E \psi \tag{6}$$
share bound state spectrum (except for the ground state for $V_{-,\ell}$), and (absolute value of) the reflection and transmission coefficients, cf. Ref. 1. Hence, we have linked all the considered potentials
$$\begin{align} 0~\sim~& V_{-,0}~\sim~ V_{+,1}~\stackrel{\text{SUSY}}{\longleftrightarrow}~ V_{-,1}\cr
~\sim~& V_{+,2}~\stackrel{\text{SUSY}}{\longleftrightarrow}~ V_{-,2}~\sim~ V_{+,3}~\stackrel{\text{SUSY}}{\longleftrightarrow}~\ldots \end{align}\tag{7}$$
to the constant potential. This explains why the sequence $(1)$ consists of reflectionless potentials for a non-negative integer $\ell\in\mathbb{N}_{0}$.
Finally, if $\ell \notin \mathbb{Z}$ is not an integer, the superpotential $(5)$ still makes sense. However, the potential $(1)$ cannot be linked via the use of supersymmetric partners and constant shifts to the trivial potential, and in fact, the potential $(1)$ is not reflectionless if $\ell \notin \mathbb{Z}$.
References:
- F. Cooper, A. Khare, & U. Sukhatme, Supersymmetry and Quantum Mechanics, Phys. Rept. 251 (1995) 267, arXiv:hep-th/9405029; Chapter 2.
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$^1$ For completeness, let us mention that there is also a sequence of reflectionless repulsive potentials. The analogues of eqs. $(1)$, $(2)$ and $(5)$ read
$$ V_{\ell}(x)~:=~\frac{(\kappa a)^2 \ell(\ell+1)}{\sinh^2 ax}, \qquad \ell~\in~\mathbb{N}_{0},\tag{1'}$$
$$V_{\pm,\ell}(x)~:=~ (\kappa a)^2\left(\ell^2 +\frac{\ell(\ell \mp 1)}{\sinh^2 ax} \right), \tag{2'} $$
$$W_{\ell}(x)~:=~\ell \kappa a \coth ax , \tag{5'} $$
respectively. For $a\to 0$, this becomes
$$ V_{\ell}(x)~:=~\frac{\kappa^2 \ell(\ell+1)}{x^2}, \qquad \ell~\in~\mathbb{N}_{0},\tag{1''}$$
$$V_{\pm,\ell}(x)~:=~ \frac{\kappa^2\ell(\ell \mp 1)}{x^2} , \tag{2''} $$
$$W_{\ell}(x)~:=~\frac{\ell \kappa}{x} ,\tag{5''} $$
respectively. Also, we have, for notational simplicity, suppressed a freedom to shift the potential profiles along the $x$-axis $x\to x-x_0$.
$^2$ For a starter, one has to assume that both the limits $\lim\limits_{x\to\pm\infty} W(x)$ exist and are finite, which holds in OP's case.
$^3$ The TISE $(6)$ can be transformed into the associated Legendre differential equation, which famously describes spherical harmonics and angular momentum states in QM with azimuthal quantum number $\ell\in\mathbb{N}_{0}$.
