Let $V \subset L^2(\mathbb{R})$ be the space of functions supported on $E$, and $W$ the space of functions whose Fourier transform is supported on $F$. (This notation differs from Jason's hint, sorry.) Note both subspaces are closed. Use the A-B theorem to show that $V \cap W = 0$, which is easy, and that $V+W$ is closed, which is a little harder. One approach to the latter is this: any function in $V+W$ can be written uniquely as $f+g$ where $f \in V$, $g \in W$. Then $V+W$ is closed iff the operators $f+g \mapsto f$ and $f+g \mapsto g$ are bounded, for then the map $f+g \mapsto (f,g)$ is an isomorphism of $V+W$ with $V \oplus W$.
To show the latter, we can use the A-B theorem to write, for $f \in V$, $g \in W$, $$\require{cancel}\begin{align*} \|f\|_{L^2(\mathbb{R})} &\le C(\cancel{\|f\|_{L^2(E^c)}} + \|\mathcal{F}(f)\|_{L^2(F^c)}) \\ &= C \|\mathcal{F}(f+g)\|_{L^2(F^c)} && \text{since $\mathcal{F}g = 0$ on $F^c$} \\ &\le C \|\mathcal{F}(f+g)\|_{L^2(\mathbb{R})} \\ &= C\|f+g\|_{L^2(\mathbb{R})} && \text{Plancherel}. \end{align*}$$
So the map $f+g \mapsto f$ is bounded, and a similar argument works for $f+g \mapsto g$.
Now that $V+W$ is closed, consider the orthogonal projections $P_V h = 1_E \cdot h$ and $P_W h = \mathcal{F}^{-1} (1_F \cdot \mathcal{F}h)$. It's a general fact, I think, that (*) when $V \cap W = 0$ and $V+W$ is closed, then for every $f \in V$, $g \in W$, there exists $h \in V+W$ with $P_V h = f$ and $P_W h = g$. This suffices for the claim, by taking $f=g_1$ and $g=\mathcal{F}^{-1} g_2$.
I don't have a reference offhand for (*), though I would bet it can be found in Halmos's A Hilbert Space Problem Book. I did come up with a proof but it's messy, surely not the one from "The Book". It's based on the fact that when $V+W$ is closed, there is a positive angle between $V$ and $W$, and using this in a roundabout way to show that $P_V + P_W$ is invertible on $V+W$.