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
$$ \tag{1'} V_{\ell}(x)~:=~\frac{(\kappa a)^2 \ell(\ell+1)}{\sinh^2 ax}, \qquad \ell~\in~\mathbb{N}_{0},$$
$$\tag{2'} V_{\pm,\ell}(x)~:=~ (\kappa a)^2\left(\ell^2 +\frac{\ell(\ell \mp 1)}{\sinh^2 ax} \right), $$
$$\tag{5'} W_{\ell}(x)~:=~\ell \kappa a \coth ax , $$
respectively. For $a\to 0$, this becomes
$$ \tag{1''} V_{\ell}(x)~:=~\frac{\kappa^2 \ell(\ell+1)}{x^2}, \qquad \ell~\in~\mathbb{N}_{0},$$
$$\tag{2''} V_{\pm,\ell}(x)~:=~ \frac{\kappa^2\ell(\ell \mp 1)}{x^2} , $$
$$\tag{5''} W_{\ell}(x)~:=~\frac{\ell \kappa}{x} , $$
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 starter, one has to assume that both the limits $\lim_{x\to\pm\infty} W(x)$ exist and are finite, which hold 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 $\ell\in\mathbb{N}_{0}$.
