= Solution
Split the free <scalar field> into its annihilation part $\phi^{(+)}$ and creation part $\phi^{(-)}$. <Normal ordering> moves every creation operator to the left of every annihilation operator, without adding the commutators produced by that rearrangement. For two fields,
$$
:\phi(x)\phi(y):=
\phi^{(-)}(x)\phi^{(-)}(y)+\phi^{(-)}(x)\phi^{(+)}(y)
+\phi^{(-)}(y)\phi^{(+)}(x)+\phi^{(+)}(x)\phi^{(+)}(y).
$$
Its vacuum expectation is zero. <Time ordering> instead places the field at the later time on the left:
$$
T\{\phi(x)\phi(y)\}
=\theta(x^0-y^0)\phi(x)\phi(y)+\theta(y^0-x^0)\phi(y)\phi(x).
$$
The equal-time convention is immaterial away from coincident singularities; these expressions are understood as <operator-valued distributions>.
The only nonzero vacuum contraction uses $a_{\mathbf p}a^\dagger_{\mathbf q}$. The oscillator <commutator> therefore gives the <Wightman function>
$$
W(x-y)=\langle0|\phi(x)\phi(y)|0\rangle
=\int\frac{d^3p}{(2\pi)^3}\frac{e^{-iE_{\mathbf p}(x^0-y^0)+i\mathbf p\cdot(\mathbf x-\mathbf y)}}{2E_{\mathbf p}}.
$$
Put $\tau=x^0-y^0$ and $\mathbf r=\mathbf x-\mathbf y$. Combining the two time orders, and reversing $\mathbf p$ in the second when necessary, gives
$$
\Delta_F(\tau,\mathbf r)
=\int\frac{d^3p}{(2\pi)^3}\frac{e^{i\mathbf p\cdot\mathbf r-iE_{\mathbf p}|\tau|}}{2E_{\mathbf p}}.
$$
To recover the four-dimensional <Fourier transform>, use the <Feynman i-epsilon prescription> in the energy variable:
$$
\int\frac{dp^0}{2\pi}\frac{i e^{-ip^0\tau}}{(p^0)^2-E_{\mathbf p}^2+i0}
=\frac{e^{-iE_{\mathbf p}|\tau|}}{2E_{\mathbf p}}.
$$
The positive-energy pole lies just below the real axis, and the negative-energy pole just above it. For $\tau>0$, close below clockwise: the residue at $E_{\mathbf p}-i0$ is $i e^{-iE_{\mathbf p}\tau}/(2E_{\mathbf p})$, and the clockwise factor $-i$ after dividing by $2\pi$ gives the required positive coefficient. For $\tau<0$, close above counterclockwise: the residue at $-E_{\mathbf p}+i0$ is $-i e^{iE_{\mathbf p}\tau}/(2E_{\mathbf p})$, and the factor $+i$ gives the other time order. Equivalently the energy contour passes above the positive-energy pole and below the negative-energy pole. Hence
$$
\boxed{\Delta_F(x-y)=\lim_{\epsilon\downarrow0}
\int\frac{d^4p}{(2\pi)^4}\frac{i e^{-ip\cdot(x-y)}}{p^2-\mu^2+i\epsilon}.}
$$
The limit is distributional. The <scalar Feynman propagator pole prescription> is essential; omitting it from the denominator must be accompanied by the specified contour.
Finally, moving the annihilation part of $\phi(x)$ past the creation part of $\phi(y)$ gives
$$
\phi(x)\phi(y)=:\phi(x)\phi(y):+W(x-y)\mathbf1.
$$
The normal-ordered two-field product is symmetric in $x,y$, because the two creation parts commute and the two annihilation parts commute. Applying the two time orders thus proves the <two-field scalar Wick identity>
$$
\boxed{T\{\phi(x)\phi(y)\}=:\phi(x)\phi(y):+\Delta_F(x-y)\mathbf1.}
$$
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