= Solution
Exhaust $\mathbb Z^2$ by finite boxes $G_n$ and let $T_n$ be a <uniform spanning tree> of $G_n$. Every finite tree satisfies the <handshaking lemma>, so
$$
\frac1{|V(G_n)|}\sum_{v\in V(G_n)}\deg_{T_n}(v)
=\frac{2(|V(G_n)|-1)}{|V(G_n)|}\longrightarrow2.
$$
Choose the root uniformly from $V(G_n)$. The proportion of roots within any fixed distance of the boundary tends to zero, and the rooted trees converge locally to the uniform spanning tree $T$ of $\mathbb Z^2$. Since every degree is at most four, expectations also converge. Translation invariance therefore gives
$$
\mathbb E[\deg_T(0)]=2.
$$
The four edges incident to $0$ have equal inclusion probability by the rotations and reflections of the square lattice. If that common probability is $r$, then $4r=\mathbb E\deg_T(0)=2$. Consequently
$$
\boxed{\mathbb P(e\in T)=\frac12}
$$
for every edge $e$ by translation invariance.
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