I'll expand on Nelson's proof a bit.
You know the Mean Value Property that the value of a harmonic function $u$ at $z$ equals the average of its values on any circle centered at $z$. We can turn this into a fact about averages over balls (which in this case are disks) just by using polar coordinates:
$$\int_{B(z,R)} u(w)\,dw = \int_0^R \int_0^{2\pi} u(z + r e^{i \theta}) r \,d\theta\,dr = \int_0^R 2 \pi u(z) r\,dr = u(z) \pi R^2.$$
Dividing both sides by $\pi R^2$, we see that $u(z)$ equals the average value of $u$ over the ball $B(z,R)$.
Now fix $z,w \in \mathbb{C}$, and consider disks $B(z,R)$, $B(w,R)$ centered at each of them. Let $L = B(z,R) \cap B(w,R)$ be the lens-shaped region lying in both disks. Let $L_z = B(z,r) \backslash L$ be the region in the $z$ disk but not the $w$ disk. (Draw a picture.) One can work out the area $A(L)$ of $L$ by a bit of multivariable calculus (or look it up on Mathworld). A bit more calculus shows that the ratio $\frac{A(L)}{\pi R^2} \to 1$ as $R \to \infty$, so that the lens occupies "most" of the disk. Since $A(L) + A(L_z) = \pi R^2$, we must therefore have $\frac{A(L_z)}{\pi R^2} \to 0$.
Now by the mean value property, for any $R$ we have
$$u(z) = \frac{1}{\pi R^2} \int_{B(z,R)} u(\zeta)\,d\zeta = \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta + \frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta.$$
For the second term, if $|u|$ is bounded by some constant $M$, we have as $R \to \infty$,
$$\left|\frac{1}{\pi R^2} \int_{L_z} u(\zeta)\,d\zeta \right| \le \frac{M A(L_z)}{\pi R^2} \to 0.$$
Thus
$$u(z) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$
But if we replace $z$ with $w$ we can make exactly the same argument (replacing $L_z$ by $L_w$ whose area is the same, by symmetry). So we also have
$$u(w) = \lim_{R \to \infty} \frac{1}{\pi R^2} \int_L u(\zeta)\,d\zeta.$$
Thus $u(z) = u(w)$.
Okay, I don't see that Ahlfors intended any particular method to be used (doesn't mean he didn't), so let's just use any method we can find, dominated convergence is the first that comes to mind.
First, for convenience, let's rotate it and consider
$$\int_{-\pi}^\pi \log\,\lvert 1 - e^{i\vartheta}\rvert \, d\vartheta,$$
so that we have the problematic point at $\vartheta = 0$.
A logarithmic singularity is integrable, since
$$ \int_\varepsilon^1 \log t\, dt = \left[t\log t -t\right]_\varepsilon^1 = -1 - \varepsilon\log\varepsilon + \varepsilon \to -1$$
for $\varepsilon \to 0$.
On the ray $r\cdot e^{i\vartheta}$ (for $\lvert \vartheta\rvert < \pi/2$), we consider
$$\delta(r) = \lvert 1- re^{i\vartheta}\rvert^2 = (1 - r\cos\vartheta)^2 + (r\sin\vartheta)^2 = 1+r^2 - 2r\cos\vartheta.$$
Since $\delta'(r) = 2(r - \cos\vartheta)$, $\delta(r)$ is minimised for $r = \cos\vartheta$, and $\delta(\cos\vartheta) = 1 - \cos^2\vartheta = \sin^2\vartheta$. That means that for any $0 < \varepsilon < \frac{\pi}{3}$ we have $\lvert \sin \vartheta\rvert \leqslant \lvert 1 - re^{i\vartheta}\rvert \leqslant 1$ for all $r\in [0,1]$ and $\lvert\vartheta\rvert \leqslant \varepsilon$, so the integrands $f_r(\vartheta) = \log\,\lvert 1 - re^{i\vartheta}\rvert$ are dominated by the integrable function $\log \frac{1}{\lvert\sin\vartheta\rvert}$ on $[-\varepsilon,\,\varepsilon]$. Outside that interval everything is bounded and convergence is uniform, so
$$\int_{-\pi}^\pi \log \,\lvert 1 - e^{i\vartheta}\rvert\,d\vartheta = \lim_{r \to 1} \int_{-\pi}^\pi \log\, \lvert 1 - re^{i\vartheta}\rvert\,d\vartheta = 0.$$
Another method that can be useful when you have a non-integrable limit, but such that the principal value of the integral exists (when you have a simple pole on the integration path, for example), is the removal of a small disk around the problematic point.
Now it would be good if I could draw a picture here, but since I can't, a description must do. Consider
$$B_\varepsilon := \mathbb{D}\setminus \overline{D_\varepsilon(1)} = \{z \in \mathbb{C} \colon \lvert z\rvert < 1, \lvert z-1\rvert > \varepsilon\}.$$
In a (simply connected) neighbourhood of $\overline{B}_\varepsilon$, the integrand is the real part of a holomorphic function ($\log (1-z)$), hence the (multiple of the) Cauchy integral
$$\int_{\partial B_\varepsilon} \frac{\log (1-z)}{z - z_0}\,dz$$
is $2\pi i \log (1-z_0)$ and vanishes for $z_0 = 0$.
The real part of that integral is then
$$\int_{\delta}^{2\pi-\delta} \log\, \lvert 1 - e^{i\vartheta}\rvert\, d\vartheta - \operatorname{Re} \int_{\partial B_\varepsilon \cap \mathbb{D}} \frac{\log (1-z)}{z}\,dz,$$
where $\delta = 2\arcsin(\varepsilon/2)$. The first integral is in the limit what we want, so we need to show that the second integral converges to $0$ for $\varepsilon \to 0$. The standard estimate does that,
$$\left\lvert \int_{\partial B_\varepsilon \cap \mathbb{D}} \frac{\log (1-z)}{z}\,dz\right\rvert \leqslant \pi\varepsilon \cdot \frac{\lvert\log \varepsilon\rvert + \pi/2}{1-\varepsilon}.$$
Best Answer
It's the average value of $\log|rz - z_0|$ over the unit circle $|z| = 1$. If $|z_0| > r$, the function is harmonic throughout the interior of the circle and is continuous on the boundary. So by the mean value theorem this average value is the value of the function at the center $z = 0$, namely $\log|z_0|$.
In the case where $|z_0| < r$, since $|re^{i\theta} - z_0| = |r - z_0e^{-i\theta}|$, your integral is the same as $${1 \over 2\pi}\int_0^{2\pi} \log|r - z_0e^{-i\theta}|\,d\theta$$ Changing variables from $\theta$ to $-\theta$, this equals $${1 \over 2\pi}\int_0^{2\pi} \log|r - z_0e^{i\theta}|\,d\theta$$ Now we have the average value of the function $\log|r - z_0z|$ over the unit circle, and since $|z_0| < r$ this function is harmonic inside the circle and continuous on the boundary. Thus the average value is that at the origin, namely $\log|r|$.
(Note as some commenters mentioned you have the values reversed.)