= Solution
Put $z=x-y$ and $d\Pi_p=d^3\mathbf p/[(2\pi)^3 2E_{\mathbf p}]$. The <Dirac adjoint> of the printed <mode expansion of a Dirac field> is
$$
\bar\psi_a(x)=\int\frac{d^3\mathbf p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}\sum_s
\left[e^{ip\cdot x}a^{s\dagger}_{\mathbf p}\bar u_a^s(\mathbf p)+e^{-ip\cdot x}b^s_{\mathbf p}\bar v_a^s(\mathbf p)\right].
$$
The <Fock vacuum> is annihilated by $a,b$. The <canonical anticommutation relations> therefore give the two non-time-ordered <vacuum expectation values>
$$
\begin{aligned}
\langle0|\bar\psi_a(x)\psi_b(y)|0\rangle&=\int d\Pi_p\,e^{-ip\cdot z}\sum_s\bar v_a^s(\mathbf p)v_b^s(\mathbf p),\\
\langle0|\psi_b(y)\bar\psi_a(x)|0\rangle&=\int d\Pi_p\,e^{ip\cdot z}\sum_s\bar u_a^s(\mathbf p)u_b^s(\mathbf p).
\end{aligned}
$$
In the first line, the only surviving contraction is $\langle b b^\dagger\rangle$; in the second it is $\langle a a^\dagger\rangle$. Both of these unordered contractions have positive coefficients.
However, <fermionic time ordering> includes a minus sign when it interchanges two odd fields:
$$
T\bar\psi_a(x)\psi_b(y)=\theta(z^0)\bar\psi_a(x)\psi_b(y)-\theta(-z^0)\psi_b(y)\bar\psi_a(x).
$$
Thus the result following from the stated <canonical anticommutation relations> and standard <time ordering> is
$$
\boxed{\begin{aligned}
C_{ab}(x,y)&\equiv\langle0|T\bar\psi_a(x)\psi_b(y)|0\rangle\\
&=\int d\Pi_p\left[\theta(z^0)e^{-ip\cdot z}\sum_s\bar v_a^s v_b^s
-\theta(-z^0)e^{ip\cdot z}\sum_s\bar u_a^s u_b^s\right].
\end{aligned}}
$$
\b[The second term printed with a plus sign is not correct for standard <fermionic time ordering>.] Its plus-plus expression would be an ungraded chronological product. The missing minus cannot be absorbed by a phase choice for $u$ or $v$, since each <spin sum> contains a spinor and its adjoint.
For comparison with the conventional <Dirac propagator>, define $S_F(w)_{ba}=\langle0|T\psi_b(w)\bar\psi_a(0)|0\rangle$ and let $\mu$ denote this free <Dirac field>'s mass. Then
$$
\boxed{C_{ab}(x,y)=-S_F(y-x)_{ba}
=-\int\frac{d^4p}{(2\pi)^4}\frac{i(\not{p}+\mu)_{ba}}{p^2-\mu^2+i0}\,e^{-ip\cdot(y-x)}.}
$$
The exchange of the spinor indices is essential: $\bar v_a v_b$ is the $(b,a)$ component of $v\bar v$, not its $(a,b)$ component.
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