I've tried solving this with the Matrix-Tree Theorem using the Laplacian matrix, and found a result of $2 \cdot (n-2)^{n-2}$.
However, I feel as if I am approaching this in the wrong way. Any hints on approaching it in a different way would be appreciated.
Your expression $2(n-2)^{n-2}$ can't be right, because when $n=3$ you are counting the spanning trees in the diamond graph $K_4-e$ and there are $8$ of them, not $2$.
Let's try just using Cayley's formula. Say the new vertex $z$ is joined to the two distinct vertices $x$, $y$ of $K_n$. A spanning tree for the new graph must contain one or both of the edges $xz$ and $yz$.
The number of spanning trees containing $xz$ but not $yz$ is equal to the number of spanning trees in $K_n$, which is $n^{n-2}$. The number of spanning trees containing $yz$ but not $xz$ is the same.
The number of spanning trees containing both $xz$ and $yz$ is equal to the number of spanning trees in the original graph $K_n$ containing the edge $xy$, and that number is $$\frac{n-1}{\binom n2}n^{n-2}=2n^{n-3}.$$
Thus the total number of spanning trees is $$2n^{n-2}+2n^{n-3}=2(n+1)n^{n-3}.$$