Modification of Riemann tensor by twisting the spin connection

Suppose one has a funfbein $e^a_{mu}$ with associated spin connection $omega^mu_{,ab}$. Writing this in terms of differential forms,
$$e_a:=e_a^{mu}mathrm{d}x^{mu}, omega_{ab}:=omega^mu_{,ab}mathrm{d}x^{mu}$$
the Riemann tensor
$$R^a_{ ,b}:=frac{1}{2}R^a_{ ,bmunu}mathrm{d}x^{mu}mathrm{d}x^{nu}$$
is given by
$$R^{a}_{ ,b}=mathrm{d}omega^a_{ ,b}+omega^a_{ ,c}landomega^c_{ ,b}.$$
All this is standard GR and I’m good with it. However, now consider a "twisted" torsionful spin connection
$$tilde{omega}_{ab}=omega_{ab}-frac{1}{4}e_fepsilon^{fabcd}F_{cd},$$
where $F_{ab}=F_{munu}e^mu_ae^nu_b$ is the graviphoton field strength. (I do not know if "twisted" is necessarily the right technical term for this.) Call the new Riemann tensor associated to $tilde{omega}$ to be $tilde{R}^a_{ ,b}$. I can show that this will be given by
begin{align}
tilde{R}^{ab}=&R^{ab}-frac{1}{4}epsilon^{fabcd}(F_{cd}mathrm{d}e_f+e_flandmathrm{d}F_{cd})
&-frac{1}{2}omega^{[a|}_{ ,e}land e_fepsilon^{f|b]ecd}F_{cd}+frac{1}{16}(F_{[cd}e_{f]})^2,
end{align}
where $[…]$ denotes antisymmetrization of indices. However, I was wondering if there was a way to rewrite this entirely in terms of curvature tensors (Riemann, Ricci, and scalar) and the field strength $F$, without explicit reference to the funfbein or spin connection?