Is it possible to have a dispersion interaction with no inductive interactions?
The first-order correction in the interaction operator is an integral over the frozen reference densities, so in a product basis $|i_A,j_B\rangle$ that would read
$$ E^{(1)} = \langle 00|V|00 \rangle $$
and as such is the electrostatic interaction. This is the same as the Coulomb interaction of the two non-interacting reference densities.

Induction is defined as the  polarization of one monomer due to the reference density of the other monomer, so must involve integrals over the virtual orbitals of the monomer being polarized in order to capture its polarization. So induction is a second-order correction and has the form
$$ E^{(2)}_{ind} = \sum_{j\neq 0} \frac{|\langle 00|V|0j \rangle|^2}{E_{00}^{(0)}-E_{0j}^{(0)}} + \sum_{i\neq 0}\frac{|\langle 00|V|i0 \rangle|^2}{E_{00}^{(0)}-E_{i0}^{(0)}}$$
where the first term represents the polarization of monomer $B$ due to the reference density of monomer $A$ and the second term is the reverse process.

What sets the dispersion energy apart from the induction energy is that it is the simultaneous polarization of both monomers, and so the integrals run over the virtual orbitals of both monomers as in
$$ 
E^{(2)}_{disp} = \sum_{i\neq 0} \sum_{j\neq 0}\frac{|\langle 00|V|ij \rangle|^2}{E_{00}^{(0)}-E_{ij}^{(0)}}
$$
Mathematically, the dispersion and induction terms are both second-order correction terms in the energy expansion, but they have been grouped in a physically-intuitive way to give two different types of interactions. There are systems where dispersion is the dominant interaction, such as rare gas dimers.

EDIT:

An example of a dispersion-dominated bound system is the neon dimer. Below is the SAPT decomposition of the binding curve at the HF-SAPT(2)/aug-cc-pVDZ level.



At the equilibrium distance, dispersion dominates the binding interaction but there is a non-zero induction contribution. I am inclined to say that although the ratio of dispersion and induction may become large, you can never truly have zero induction. Some naive mathematical reasoning follows.

We need $ \langle 00|\frac{1}{r_{12}}|0i\rangle, i\neq 0$ to be zero in a Gaussian product basis. At a long-enough distance we can expand the Coulomb operator in a multipole expansion to give something like
$$ 
\frac{1}{r_{12}} = \alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)+...
$$
in a one-dimensional model picture. I've also folded all the coefficients into the constants $\alpha,\beta,...,etc.$ The important thing here is the symmetry of the integrals. So, lets take $i=1$ and see what happens...
$$
\langle 00|\alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)|01\rangle = \langle 00|\alpha x_1x_2|01\rangle + \langle 00|\beta x_1x_2^2|01\rangle-\langle 00|\beta x_1^2x_2|01\rangle+ ...
$$
The first two integrals are zero because $\langle 0|x_1|0\rangle=0$. The third integral is non-zero and survives in the sum. This can happen any time the operator in monomer $A$ is an even function. As $i$ gets larger, higher-order terms in the expansion contribute to the induction energy. So mathematically, there is a tiny but non-zero induction contribution.