For $m = 2$ you can simply use Kelvin transform to exchange the roles of $y_0$ and $\infty$, as pointed out by Alexandre Eremenko. Thus, we assume that $m \geqslant 3$.
Let $u$ be a superharmonic function in a neighbourhood of a compact set $K$ such that the complement of $K$ is connected. In your case $u$ is equal in $K$ to the Newtonian potential kernel $U(x)$. Let $y_0$ be a fixed point in $\mathbb R^m \setminus K$. By changing $x$ to $y_0 + x$, with no loss of generality we may assume that $y_0 = 0$.
As you write in the other question, by Theorem 6.10.1 in [Armitage, Gardiner, Classical Potential Theory], there is a superharmonic function $v$ in all of $\mathbb R^m$ such that $u = v$ in $K$ and $$v = \alpha + \beta U \qquad \text{ in } \mathbb R^m \setminus B(0, R)$$ for some $R > 0$, $\alpha \in \mathbb R$ and $\beta \geqslant 0$.
Now let $w(x)$ be the Green function of $\mathbb R^m \setminus K$ with pole at $0$ (that is, $w = U - h$ for a harmonic function $h$ in $\mathbb R^m \setminus K$ such that $h = U$ on $\partial K$ — up to irregular boundary points, $h \ge 0$, and $h \to 0$ at infinity). Consider $$\overline u = v - \gamma w$$ for $\gamma$ large enough, so that $$\gamma w \geqslant \beta U \qquad \text{ on } \partial B(0, R), \text{ and hence on all of } \mathbb R^n \setminus B(0, R)$$ (the latter claim follows from the former by the maximum principle applied to $\gamma w - \beta U$). Clearly, $\overline u = v = u$ in $K$. Furthermore, both $v$ and $-w$ are superharmonic in $\mathbb R^m \setminus \{y_0\}$, and so $\overline u$ is superharmonic in $\mathbb R^m \setminus \{y_0\}$, too. Finally, $$\overline u = (\alpha + \beta U) - \gamma w \leqslant \alpha \qquad \text{ in } \mathbb R^m \setminus B(0, R).$$ Since $\overline u \to \alpha$ at infinity, the mean value of $\overline u$ over any (large enough) sphere does not exceed $\alpha$. Thus, $\overline u$ is the desired extension of $u$.
Edit: Here are some details on the $m = 2$ case.
Let $u$ be a superharmonic function in a neighbourhood of a compact set 𝐾 such that the complement of 𝐾 is connected, and let $y_0 \in \mathbb R^2 \setminus K$. Again, with no loss of generality we assume that $y_0 = 0$.
For $x \ne 0$ let $x^*$ be the image of $x$ under the inversion with respect to the unit sphere: $$x^* = |x|^{-2} x.$$ Denote $$K^* = \{x^* : x \in K\}$$ and $$u^*(x^*) = |x|^{2 - m} u(x)$$ denote the Kelvin transform of $u$. Then it is a classical result that $u^*$ is superharmonic in a neigbourhood of $K^*$.
Let $v^*$ be the superharmonic extension of $u^*$ to all of $\mathbb R^m$, as in the first part of the answer (note that the complement of $K^*$ is connected), and let $$v(x) = v^*(x^*)$$ be the Kelvin transform of $v^*$. Then $v^*$ is superharmonic in $\mathbb R^m \setminus \{0\}$. Furthermore, if we set $v(\infty) = v^*(0)$ and if the dimension is 2 ($m = 2$), then $$v(\infty) = v^*(0) \geqslant \frac{r}{2\pi} \int_{\partial B(0, 1/r)} v^* = \frac{1}{2\pi r} \int_{\partial B(0, r)} v \tag{$\star$}$$ for $r$ large enough, which (almost) shows that $v$ is superharmonic at infinity. Almost, because we only consider balls centred at the origin. For the general case, one has the same argument, but this time involving the Poisson kernel: $$v(\infty) = v^*(0) \geqslant \int_{\partial B(x_0', r')} v^*(y) P_{B(x_0', r')}(0, y) dy = \frac{1}{2\pi r} \int_{\partial B(x_0, r)} v , $$ where $\partial B(x_0', r')$ is the image of $\partial B(x_0, r)$ under the inversion. (The calculations get a bit messy here, but they are rather standard.)
Note: in dimensions $m \geqslant 3$, one can follow the same argument, but the scaling in ($\star$) is wrong.
