Question on the Prize-Collecting TSP's ratio related to inapprox. of general TSP
Edited may answer, I hope it becomes clearer now (or I find out what I did not understand):

Your process is as follows: If you have an edge $e = \{u,v\}$ with weight $t$ you replace $e$ by a path of length two with edges $\{u,z_e\}$ and $\{z_e,v\}$. $z_e$ is a new node and each of the two new edges get weight $t/2$. For each edge, we introduce a new node.

So far, we do not have a metric instance, because there is no edge between $u$ and $v$ so far and the distance between $u$ and $v$ is infinite. Therefore, you take the shortest path metric to get these distances. (I suppose that this is what you mean by "having Euclidean distances equal to that". Otherwise, you should clarify this.)

My point is that after adding distances according to the shortest path metric, the distance between $u$ and $v$ might not be $t$ anymore, so you do not get your original instance back:
Consider a triangle with edge lengths $1$, $1$ and $t$. Let $u$ and $v$ be the nodes of the edge with weight $t$. Apply your construction. The shortest path between $u$ and $v$ is $2$ (it takes the four edges with weight $1/2$) and not $t$.

Btw: Euclidean distance should just be distance in the question.