The Vorticity Evolution in 2D Cartesian Coordinates, assuming incompressibility, is as follows: $$ \frac{\partial \omega}{\partial t} = \nu \left( \frac{\partial^2 \omega}{\partial x^2} + \frac{\partial^2 \omega}{\partial y^2} \right) - \left( u \frac{\partial \omega}{\partial x} + v \frac{\partial \omega}{\partial y} \right) $$ By using discrete spectral methods, assuming periodic boundary conditions, where $\Omega_{pq} = \mathrm{fft2}(\omega_{ij})$, our equation becomes: $$ \frac{\partial \Omega_{pq}}{\partial t} = \nu \left( \frac{\partial^2 \Omega_{pq}}{\partial x^2} + \frac{\partial^2 \Omega_{pq}}{\partial y^2} \right) - \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) $$
If we replace the derivative operator with spectral derivatives ($ \partial / \partial x = + \underline{i} k_p$), we obtain the following: $$ \frac{\partial \Omega_{pq}}{\partial t} = - \nu \left( k_p^2 + k_q^2 \right) \Omega_{pq} - \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) $$ I have been given the following equation, which claims to be equivalent to those above, where $\Xi_{pq} = \exp \left[ \nu \left( k_p^2 + k_q^2 \right) t \right]$ is an integrating factor: $$ \frac{\partial}{\partial t} \Big[ \Xi_{pq} \Omega_{pq} \Big] = -\underline{i} \Xi_{pq} \Big[ k_p \cdot \mathrm{fft2} \left(u_{ij} \omega_{ij} \right) + k_q \cdot \mathrm{fft2} \left( v_{ij} \omega_{ij} \right) \Big]$$
Unfortunately, I don't understand how these are equivalent, though a numerical simulation with Python seems to show agreement. To aid those that don't recognize what these variables represent and the relations between them, the following relations are true by definition: $$ \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = - \omega \quad \to \quad \Psi_{pq} = \frac{\Omega_{pq}}{k_p^2 + k_q^2} $$ $$ u = +\frac{\partial \psi}{\partial y} \quad \to \quad U_{pq} = + \underline{i} k_q \Psi_\mathrm{pq} = + \frac{\underline{i} k_q \Omega_\mathrm{pq}}{k_p^2 + k_q^2} $$ $$ v_\mathrm{ij} = - \frac{\partial \psi_\mathrm{ij}}{\partial x} \quad \to \quad V_\mathrm{pq} = - \underline{i} k_p \Psi_\mathrm{pq} = - \frac{\underline{i} k_p \Omega_\mathrm{pq}}{k_p^2 + k_q^2} $$ $$ \nabla \times \vec{u} = \Big \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \Big \rangle \times \langle u, v \rangle = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = \omega $$ $$ \nabla \cdot \vec{u} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$$
Starting with the following: $$ \frac{\partial \Omega_{pq}}{\partial t} = - \nu \left( k_p^2 + k_q^2 \right) \Omega_{pq} - \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) $$ $$ \implies \frac{\partial \Omega_{pq}}{\partial t} + \nu \left( k_p^2 + k_q^2 \right) \Omega_{pq} = - \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) $$ It can be seen that $ \nu \left( k_p^2 + k_q^2 \right) \exp \left( \nu \left( k_p^2 + k_q^2 \right) t \right) = \frac{d}{dt} \left[ \exp \left( \nu \left( k_p^2 + k_q^2 \right) t \right) \right] $ by the Chain Rule, and as a result: $$ \implies \frac{\partial}{\partial t} \Big[ \exp \left[ \nu \left( k_p^2 + k_q^2 \right) t \right] \cdot \Omega_{pq} \Big] = - \exp \left[ \nu \left( k_p^2 + k_q^2 \right) t \right] \cdot \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right)$$ $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \Xi_{pq} \cdot \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) \;\: \mathrm{where} \; \: \Xi_{pq} = \exp \left[ \nu \left( k_p^2 + k_q^2 \right) t \right] $$ Since the Fourier Transform is linear, i.e. $\mathrm{fft2}[a + b] = \mathrm{fft2}[a] + \mathrm{fft2}[b]$, we can seperate the RHS: $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \Xi_{pq} \cdot \mathrm{fft2} \left( u_{ij} \frac{\partial \omega_{ij}}{\partial x} \right) - \Xi_{pq} \cdot \mathrm{fft2} \left( v_{ij} \frac{\partial \omega_{ij}}{\partial y} \right) $$ One might be tempted to expand out the spectral derivatives for each $\omega_{ij}$ derivative, but this is invalid due to the non-linear terms the separated fourier transforms are acting on. As a result, we have to use one of the initial assumptions of incompressibility, where: $$\nabla \cdot \vec{u}_{ij} = 0 \to \frac{\partial u_{ij}}{\partial x} + \frac{\partial v_{ij}}{\partial y} = 0$$ $$ u_{ij} \frac{\partial \omega_{ij}}{\partial x} = \frac{\partial}{\partial x} \left[ u_{ij} \omega_{ij} \right] - \frac{\partial u_{ij}}{\partial x} \omega_{ij} \quad \mathrm{and} \quad v_{ij} \frac{\partial \omega_{ij}}{\partial y} = \frac{\partial}{\partial y} \left[ v_{ij} \omega_{ij} \right] - \frac{\partial v_{ij}}{\partial y} \omega_{ij} $$ $$ \implies u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} = \frac{\partial}{\partial x} \left[ u_{ij} \omega_{ij} \right] + \frac{\partial}{\partial y} \left[ v_{ij} \omega_{ij} \right] - \omega_{ij} \left[ \frac{\partial u_{ij}}{\partial x} + \frac{\partial v_{ij}}{\partial y} \right] $$ $$ \implies u_{ij} \frac{\partial \omega_{ij}}{\partial x} + v_{ij} \frac{\partial \omega_{ij}}{\partial y} = \frac{\partial}{\partial x} \left[ u_{ij} \omega_{ij} \right] + \frac{\partial}{\partial y} \left[ v_{ij} \omega_{ij} \right] $$ Going back to the actual equation: $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \Xi_{pq} \cdot \mathrm{fft2} \left( \frac{\partial}{\partial x} \left[ u_{ij} \omega_{ij} \right] + \frac{\partial}{\partial y} \left[ v_{ij} \omega_{ij} \right] \right) $$ $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \Xi_{pq} \cdot \mathrm{fft2} \left( \frac{\partial}{\partial x} \left[ u_{ij} \omega_{ij} \right] \right) - \Xi_{pq} \cdot \mathrm{fft2} \left(\frac{\partial}{\partial y} \left[ v_{ij} \omega_{ij} \right] \right) $$ $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \Xi_{pq} \cdot \underline{i} k_p \mathrm{fft2} \left( u_{ij} \omega_{ij} \right) - \Xi_{pq} \cdot \underline{i} k_q \mathrm{fft2} \left(v_{ij} \omega_{ij}\right) $$ And now we arrive at the originally claimed equivalent form: $$ \implies \frac{\partial}{\partial t} \Big[ \Xi_{pq} \cdot \Omega_{pq}\Big] = - \underline{i} \Xi_{pq} \cdot \Big[ k_p \cdot \mathrm{fft2} \left( u_{ij} \omega_{ij} \right) + k_q \cdot \mathrm{fft2} \left(v_{ij} \omega_{ij}\right) \Big] \;\: \square $$
Thank you @MatthewCassell for your help in the comments.