$\require{cancel}$I) OP is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$
$$ S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|}, $$
$$ T~=~\frac{i}{2} \bar{\psi}
\stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad
V~=~ \alpha j^a \eta_{ab} j^b,
\qquad j^a~:=~ \bar{\psi} \gamma^a\psi,\qquad \bar{\psi}~:=~\psi^{\dagger}\gamma^0,$$
$$ \bar{\psi}\stackrel{\leftrightarrow}{\cancel{\nabla}}\psi
~:=~ \bar{\psi}\stackrel{\leftrightarrow}{\cancel{\partial}}\psi
+\frac{1}{2} \omega_{c, ab}~\gamma^{cab}\psi
~=~\bar{\psi}\left[\gamma^c\stackrel{\rightarrow}{\nabla_c}
-\stackrel{\leftarrow}{\nabla_c}\gamma^c\right]\psi, $$
$$\stackrel{\rightarrow}{\nabla_c}\psi
~:=~ \stackrel{\rightarrow}{\partial_c}\psi +\frac{1}{4} \omega_{c, ab}~\gamma^{ab}\psi, \qquad
\bar{\psi}\stackrel{\leftarrow}{\nabla_c}
~:=~ \bar{\psi}\stackrel{\leftarrow}{\partial_c} -\frac{1}{4} \bar{\psi}~\gamma^{ab}\omega_{c, ab},$$
$$\stackrel{\leftrightarrow}{\cancel{\partial}} ~:=~ \gamma^c\stackrel{\rightarrow}{\partial_c} - \stackrel{\leftarrow}{\partial_c}\gamma^c, \qquad
\stackrel{\rightarrow}{\partial_c}~:=~E^{\mu}{}_c \stackrel{\rightarrow}{\partial_{\mu}}, \qquad
\stackrel{\leftarrow}{\partial_c}~:=~ \stackrel{\leftarrow}{\partial_{\mu}}E^{\mu}{}_c,$$
$$
\partial_{\mu}~:=~\frac{\partial}{\partial x^{\mu}}, \qquad \gamma^{ab}~:=~\frac{1}{2}[\gamma^a,\gamma^b], \qquad
\gamma^{abc}~:=~\frac{1}{2}\{\gamma^a,\gamma^{bc}\}_+.
\tag{1} $$
II) The main point is that in order to write down a covariant kinetic term for a Dirac fermion in curved spacetime, we should use a covariant derivative $\nabla_{\mu}\psi$ of a spinor $\psi$, and hence we need a spin connection $\omega_{\mu}{}^a{}_b$. In turn, we need a vielbein
$$g_{\mu\nu}~=~e^a{}_{\mu} ~\eta_{ab}~e^b{}_{\nu}, \qquad
e^a{}_{\mu}~ E^{\mu}{}_b~=~\delta^a_b, \qquad
E^{\mu}{}_a~e^a{}_{\nu}~=~\delta^{\mu}_{\nu}, \tag{2} $$
which (we for simplicity will assume) is covariantly conserved
$$0~=~(\nabla_{\mu}e)^{a}{}_{\nu}~=~\partial_{\mu}e^{a}{}_{\nu} +\omega_{\mu}{}^a{}_b ~e^b{}_{\nu}- e^a{}_{\lambda}~\Gamma_{\mu\nu}^{\lambda}.\tag{3} $$
Hence the spin connection is complete determined
$$ 2\omega_{\mu, ab}~=~2\left(-\partial_{\mu}e_{a\nu}
+e_{a\lambda}~\Gamma_{\mu\nu}^{\lambda}\right)E^{\nu}{}_b
~=~-\left(\partial_{\mu}e_{a\nu}
+\partial_a g_{\mu\nu}\right)E^{\nu}{}_b -(a\leftrightarrow b)$$
$$ ~=~-\partial_{\mu}e_{a\nu}~E^{\nu}{}_b-\partial_a e_{b\mu} + g_{\mu\nu}~\partial_a E^{\nu}{}_b -(a\leftrightarrow b),\tag{4} $$
and
$$ 2\omega_{c, ab}~:=~2E^{\mu}{}_c~\omega_{\mu, ab}
~=~-f_{cab}-f_{abc}-f_{acb}-(a\leftrightarrow b), \tag{5}$$
where we defined
$$f_{abc}~:=~\partial_a e_{b\nu}~E^{\nu}{}_c . \tag{6}$$
III) The kinetic term becomes
$$ T~=~\frac{i}{2}\bar{\psi}\stackrel{\leftrightarrow}{\cancel{\nabla}}\psi
~=~\frac{i}{2}\bar{\psi}\stackrel{\leftrightarrow}{\cancel{\partial}}\psi -\frac{i}{4}\bar{\psi}~f_{abc}~\gamma^{cab}~\psi $$
$$ ~=~\frac{i}{2}\bar{\psi}\left[\gamma^c~E^{\mu}{}_c\stackrel{\rightarrow}{\partial_{\mu}} -\stackrel{\leftarrow}{\partial_{\mu}}E^{\mu}{}_c~\gamma^c
\right]\psi
-\frac{i}{4}\bar{\psi}~E^{\mu}{}_a~\partial_{\mu} e_{b\nu}~E^{\nu}{}_c
~\gamma^{cab}~\psi. \tag{7}$$
IV) The natural generalization of the Hilbert SEM tensor
$$T^{\mu\nu}~=~- \frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}, \qquad
T_{\mu\nu}~=~\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}},\qquad(\leftarrow\text{Not applicable!})\qquad
\tag{8} $$
to fermions is given by the formula
$$T^{\mu\nu}~=~-\frac{E^{\mu c}}{2e}\frac{\delta S}{\delta e^c{}_{\nu}}+(\mu\leftrightarrow \nu), \qquad T_{\mu\nu}~=~\frac{e_{c\mu}}{2e}\frac{\delta S}{\delta E^{\nu}{}_c}+(\mu\leftrightarrow \nu).\tag{9}$$
Formula (9) reduces to the standard Hilbert SEM tensor (8) if the action only depends on the vielbein through the metric (2). However formula (9) is more general and is necessary in the case of fermions in curved spacetime.
V) The Hilbert SEM tensor with flat indices then becomes
$$T_{cd}~:=~ E^{\mu}{}_c~ T_{\mu\nu} ~E^{\nu}{}_d~\stackrel{(9)}{=}~-\frac{e_{c\nu}}{2e}\frac{\delta S}{\delta e^d{}_{\nu}} +(c\leftrightarrow d)
~=~\frac{E^{\nu}{}_c}{2e}\frac{\delta S}{\delta E^{\nu d}} +(c\leftrightarrow d)$$
$$~\stackrel{(7)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\partial_d} -\stackrel{\leftarrow}{\partial_c}\gamma_d
+\frac{1}{2} (f_{cba}-f_{abc}-f_{acb})~\gamma_d{}^{ab} \right]\psi
-\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d)$$
$$~\stackrel{(5)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\partial_d} -\stackrel{\leftarrow}{\partial_c}\gamma_d
+\frac{1}{2} \omega_{c,ab}~\gamma_d{}^{ab} \right]\psi
-\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d)$$
$$~\stackrel{(1)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\nabla_d} -\stackrel{\leftarrow}{\nabla_c}\gamma_d\right]\psi
-\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d).
\tag{10} $$
Eq. (10) is the formula for the (generalized) Hilbert SEM tensor of a Dirac fermion in curved spacetime. This is the appropriate matter source term in the EFE, cf. OP's title question (v3). For further details, see also my Phys.SE answers here and here.
--
$^1$ One may show that the Lagrangian density (1) is real using
$$ (\gamma^a)^{\dagger}~=~ \gamma^0\gamma^a\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{11} $$
Conventions: In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra
$$\{\gamma^a,\gamma^b\}_+~=~2\eta^{ab}{\bf 1}.\tag{12}$$
Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.
Let's say, $\Psi$ is a solution of Dirac equation, that is, $$(i\gamma^{\mu}\partial_{\mu}-m)\Psi=0.$$
Multiplying by $\gamma^5$ and using $\gamma^5\gamma^{\mu}=-\gamma^{\mu}\gamma^5$,
$$(i\gamma^{\mu}\partial_{\mu}+m)\gamma^5\Psi=0.$$
Thus, $\gamma^5\Psi$ is also a solution with mass $-m$. The two solutions correspond to the two factors of $E^2=p^2+m^2$. Since Dirac equation is linear, the linear combinations of its solutions: $\psi_L=\frac{1}{2}(1-\gamma^5)\Psi$ and $\psi_R=\frac{1}{2}(1+\gamma^5)\Psi$, will also be solutions.
All this basically means that Dirac equation describes two solutions and depending upon the choice of basis (for $\gamma$-matrices) these solutions could be interpreted as particle and anti-particle (in Dirac basis) or two-component left- and right- handed Weyl spinors (in chiral basis). These components separately are solutions to the Klein-Gordon equation.
EDIT: Notice that in the last expression in the question, there is no $\gamma$-matrix, so each component satisfies the K-G equation individually. As per the disappearance of the $\pm$ sign, I think it has to do with the fact that EOM for both $\Psi$ and $\bar{\Psi}$ (Dirac equation and its adjoint, respectively) can be obtained from the same Lagrangian by varying it w.r.t. to $\bar{\Psi}$ and $\Psi$, respectively. So, in that sense, it doesn't matter which one you begin with (the Dirac Lagrangian with the minus sign or its adjoint with the plus sign). They carry the same information; they're just adjoint of each other!
For detailed explanation, go here.
A similar thread is this one.
Best Answer
As is wellknown, the EFE is a PDE for the metric $g_{\mu\nu}$ / vielbein $e^a{}_{\mu}$, which plays the role of the dynamical fields of GR.
OP is apparently pondering the following question.
Answer: No, this needs not be the case.
Examples:
OP is considering matter composed of Dirac fermions. The vielbein generalization of the Hilbert SEM tensor does depend on the spin connection, cf. my Phys.SE answer here.
Already the Maxwell SEM tensor in curved space depends on the metric $g_{\mu\nu}$.
A similar situation takes place in scalar QED. Here the EFEs are replaced with Maxwell's equations. The dynamical fields are now $A_{\mu}$. One may show that the source term (the electric $4$-current $j^{\mu}$) in this case depends on $A_{\mu}$.