Stress-energy tensor for a fermionic Lagrangian in curved spacetime - which one appears in the EFE?

$\require{cancel}$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.

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$^1$ 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.

As @Holographer has mentioned in a comment, the correct formula for the stress-tensor that enters the EFE is $$ T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{ \delta g^{\mu\nu} } $$ whereas what you are computing is the canonical stress energy tensor. However, there is a subtle relation between the two, which I will elaborate upon here.

Apart from a theory that contains only scalars, the canonical stress tensor is never the one that enters the EFE. This is because, in general, the canonical stress tensor is not symmetric and therefore cannot possibly be the same stress tensor that enters in the EFE. For instance the canonical stress tensor for electromagnetism is $$ (T^{EM}_{\mu\nu})_{\text{canonical}} = F^\rho{}_\mu \partial_\nu A_\rho + \frac{1}{4} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} $$ which is not only not symmetric, but also not gauge invariant. PS - The non-symmetry is due to the spin of the field involved and is closely related to the angular momentum tensor.

However, there is an ambiguity in the construction of the stress tensor (the ambiguity does not change the conserved charges which are physical quantities). This ambiguity allows construction of an improved stress tensor (often known as the Belinfante tensor) that is symmetric and conserved. It is this improved tensor that enters the EFE. (ref. this book)

To see the equivalence, let us recall the standard construction of the stress-tensor. Consider a coordinate transformation $$ x^\mu \to x^\mu + a^\mu (x) $$ Since the original Lagrangian is invariant under translations (where $a^\mu$ is constant), the change in the action under such a coordinate transformation is $$ \delta S = \int d^d x \sqrt{-g} \nabla_\mu a_\nu T_B^{\mu\nu} $$ Now, if the stress-tensor is symmetric then we can write $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \left( \nabla_\mu a_\nu + \nabla_\nu a_\mu \right) T_B^{\mu\nu} $$ Note that the term in the parenthesis is precisely the change in the metric under the coordinate transformation. Thus, $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \delta g_{\mu\nu} T_B^{\mu\nu} \implies \frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}} = T_B^{\mu\nu} $$

Thus, we see that the symmetric Belinfante stress tensor is precisely the gravitational stress tensor. Note of course that what I've said holds specifically in a Minkowskian background, since the construction of $T_B^{\mu\nu}$ assumes Lorentz invariance.