Let $B$ denote the $n\times n$ matrix $I_x-A$. Let $[n]=\{1,2,\ldots,n\}$.
For $I,J\subset [n]$ with the same number of elements, let $D_{I,J}$ denote the determinant of the matrix resulting from $B$ by deleting the rows indexed by the elements of $I$ and the columns indexed by the elements of $J$.

Assuming $i\neq j$, we clearly have $P=D_{\varnothing,\varnothing}$, $P_i=D_{\{i\},\{i\}}$, $P_j=D_{\{j\},\{j\}}$, and $P_{ij}=D_{\{i,j\},\{i,j\}}$.
By the <a href="https://en.wikipedia.org/wiki/Dodgson_condensation">Dodgson condensation identity</a>,
$$
D_{\{i\},\{i\}}D_{\{j\},\{j\}}-D_{\{i\},\{j\}}D_{\{j\},\{i\}}=
D_{\{i,j\},\{i,j\}}D_{\varnothing,\varnothing}\ .
$$
Since $B$ is symmetric $D_{\{i\},\{j\}}=D_{\{j\},\{i\}}$.
As a result
$$
\Delta_{ij}=(D_{\{i\},\{j\}})^2\ .
$$

In the tree case, the relation between $\Phi_{T-[v_i,v_j]}$ and $D_{\{i\},\{j\}}$ can be worked out using Theorem 1 of my article <a href="https://www.sciencedirect.com/science/article/pii/S0196885803001465">"The Grassmann–Berezin calculus and theorems of the matrix-tree type"</a> in Adv. Appl. Math. 2004. For those without access to the journal, the preprint versions is <a href="https://arxiv.org/abs/math/0306396">here</a>.