Is it just a coincidence that bi-vectors are used both for
electromagnetism and rotors?
Yes and No.
First answer: Yes, its just a coincidence. The bi-vector $B$ as in the spinor
$$
\psi =\sqrt{\rho}e^{Ib/2}e^{-B/2} \space \space\space\space(1)
$$
is not related to the electromagnetism $F$. It is just a coincidence.
The electromagnetism $F$ is basically a two-form in terms of differential geometry
$$
F= F_{\mu\nu} \ dx^\mu\wedge dx^\nu
$$
thus $F_{\mu\nu}$ transforms as a "bi-vector" under diffeomorphism transformation. On the other hand, the spinor $\psi$ is a zero-form (scalar under diffeomorphism) and hence invariant under diffeomorphism transformation. Therefore, the bi-vector $B$ is in this sense unrelated to $F$.
The bi-vector $B$ is actually related to the local Lorentz transformation (as in Lorentz gauge theory of gravity) which is a separate transformation other than the diffeomorphism transformation. Therefore, instead of related to electromagnetism, bi-vector $B$ is related to gravity via the bi-vector-valued spin connection one-form $$\omega = \omega_\mu^{ab}\gamma_a\gamma_b \ dx^\mu$$ and the bi-vector-valued Riemann curvature two-form $$R = R_{\mu\nu}^{ab}\gamma_a\gamma_b\ dx^\mu\wedge dx^\nu$$ of Lorentz gauge gravity.
Interestingly, there is indeed something within $\psi$ that IS truly related to electromagnetism $F$, which is the internal electromagnetism $U(1)$ gauge transformation related to the pseudoscalar $e^{Ib/2}$ as in equation (1) above. In this sense, the electromagnetism $F$ is more accurately described as a pseudoscalar $I=\gamma_0\gamma_1\gamma_2\gamma_3$-valued two-form
$$
F= F_{\mu\nu} I \ dx^\mu\wedge dx^\nu = F_{\mu\nu} \gamma_0\gamma_1\gamma_2\gamma_3 \ dx^\mu\wedge dx^\nu
$$
In other words, in terms of geometric algebra, $F$ is indeed a four-vector, rather than a bi-vector.
Second answer: No, its not a coincidence. Note that the previous argument of Yes answer is in the context of general covariant spacetime. However, when we talk about special relativity and the Minkowskian spacetime, the bi-vector $B$ as in the spinor $\psi$ IS somehow related to the electromagnetism $F$.
Why? The Minkowskian spacetime specific tetrad/viervbein one-form
$$
e = e_\mu^a \gamma_a\ dx^\mu = \delta_\mu^a \gamma_a dx^\mu = \gamma_\mu dx^\mu
$$
acts as a soldering form gluing together the local Lorentz index $a$ and the external differential form index $\mu$. By the nature of Minkowskian spacetime, the internal frame $\gamma_a$ rotation and external coordinate $x^\mu$ rotation are inextricably tied together by the tetrads/vierbeins, which makes the spinor $\psi$ bi-vector B related to the electromagnetism $F$ which can be effectively treated as bi-vector
$$
F = F_{\mu\nu} \gamma^\mu\gamma^\nu
$$
under the global Lorentz transformation.
