Can any harmonic function on $\{z:0<|z|<1\}$ be extended to $z=0$? [duplicate]

Yes, you can extend $u$ to a harmonic function $U$ defined on the whole disk $D=\{z\in \mathbb C:0 < |z| < 1\}$.

Take a circle of radius $0\lt r\lt 1$. The key to the problem is the remark that if the harmonic extension $U$ exists, it will be given for $\mid z\mid \lt r$ by the Poisson formula $$U(z) =\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta})P(r,\theta,z) d\theta$$ where $$ P(r,\theta,z)=\frac{r^2-\mid z\mid }{\mid re^{i\theta}-z\mid}^2 $$ is the Poisson kernel.
Conversely, $U$ defined by the formula above is harmonic on $\lbrace z\in \mathbb C:0 < |z| < r\rbrace$ : this is known as the solution to Dirichlet's problem by Poisson's integral formula.
That $U$ coincides with $u$ on $\lbrace z\in \mathbb C:0 < |z| < r\rbrace$ and thus harmonically extends $u$ through $0$ necessitates a little approximation argument that you will find on page 33 of this freely available book on harmonic functions.

Let $M = \sup_{0<|z|<1} |u(z)|$. By the hypothesis, $M$ is finite.

Now note that if $z \neq 0$, then for all $0 < r < |z|$ and $0 < \epsilon < \min\left(r, \frac{1}{2}|z|\right)$, we have

$$\left|\oint_{|\zeta|=r}\frac{u(\zeta)}{\zeta-z}\;d\zeta\right| = \left|\oint_{|\zeta|=\epsilon}\frac{u(\zeta)}{\zeta-z}\;d\zeta\right| \leq \oint_{|\zeta|=\epsilon}\left|\frac{u(\zeta)}{\zeta-z}\right|\;\left|d\zeta\right| \leq \oint_{|\zeta|=\epsilon}\frac{M}{|z|/2}\;\left|d\zeta\right| = \frac{4\pi M}{|z|}\epsilon,$$

which implies, by taking $\epsilon \to 0$, that

$$ \oint_{|\zeta|=r}\frac{u(\zeta)}{\zeta-z}\;d\zeta = 0. $$

Thus if we fix $0 < R < 1$ and define

$$\tilde{u}(z) = \frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{u(\zeta)}{\zeta-z}\;d\zeta$$

for $|z| < R$, then by a simple application of Cauchy integration formula, for all $0 < |z| < R$ we have

$$ \begin{align*} \tilde{u}(z) = \frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{u(\zeta)}{\zeta-z}\;d\zeta &= \frac{1}{2\pi i}\oint_{\partial (B_R \setminus B_{|z|/3})}\frac{u(\zeta)}{\zeta-z}\;d\zeta + \frac{1}{2\pi i}\oint_{|\zeta|=|z|/3}\frac{u(\zeta)}{\zeta-z}\;d\zeta \\ &= u(z) + 0 = u(z). \end{align*}$$

Now since $\tilde{u}(z)$ is analytic on $|z| < R$, by gluing $u$ and $\tilde{u}$ we obtain an analytic function on the unit disk which coincides with $u$ on the punctuated unit disk.