I think a good way to see this is to simplify the stage a little. Imagine we want to solve the Schrödinger equation for a Hamiltonian $H$. For this we take a very simple basis, namely just two real-valued functions $\{f_1, f_2\}$. If we variationally optimise the trial wavefunction from this basis, we are presented with an approximation to the ground state in the linear combination $ \Psi = c_1 f_1 + c_2 f_2$
and some corresponding approximate energy $E$. Since we are variational, we have
$$ 0 = \frac{dE}{dc_1} = 2 \left\langle \Psi \middle| H \frac{d\Psi}{dc_1} \right\rangle = 2 \left\langle \Psi \middle| H f_1 \right\rangle$$
and similarly $0 = \left\langle \Psi \middle| H f_2 \right\rangle$.
Now the derivative of the energy wrt. positions can be written as
\begin{equation}
\frac{dE}{d\mathbf{R}}=\left\langle\Psi \middle|\frac{dH}{d\mathbf{R}}\Psi\right\rangle
+ 2 \left\langle \Psi \middle| H \frac{d\Psi}{d\mathbf{R}} \right\rangle
\end{equation}
with the second term being the Pulay forces of interest for us. Consider as an example the derivative wrt. $R_1$. Its Pulay term is
$$ \left\langle \Psi \middle| H \left( c_1 \frac{df_1}{dR_1} + c_2 \frac{df_2}{dR_1} \right)\right\rangle. $$
When is this term zero? Either when both the derivative of $f_1$ and $f_2$ wrt. $R_1$ are zero, i.e. if the basis functions by itself is independent of atomic positions. This is the case for example for plane waves or generally all basis sets which are not atom-centred. The other option is if the derivatives $\frac{df_1}{dR_1}$ and $\frac{df_2}{dR_1}$ are itself basis functions or can be exactly represented by the basis. To see what happens then, assume this was the case. We could write
$$ \frac{df_1}{dR_1} = k_{11} f_1 + k_{12} f_2 \quad\text{and}\quad \frac{df_2}{dR_1} = k_{21} f_1 + k_{22} f_2$$
for appropriate constants and obtain
\begin{align}
&\hspace{-30pt}\left\langle \Psi \middle| H \left[ (c_1k_{11} + c_2k_{21}) f_1 + (c_1k_{12} + c_2k_{22}) f_2 \right]\right\rangle \\
&= (c_1k_{11} + c_2k_{21}) \left\langle \Psi \middle| H f_1 \right\rangle + (c_1k_{12} + c_2k_{22}) \left\langle \Psi \middle| H f_2 \right\rangle \\
&= 0
\end{align}
due to the first expressions we derived.
A complete basis set is just a special case of our argument where by definition all derivatives of basis functions can be represented by the basis itself, thus giving net zero Pulay forces.