Can any Steiner Tree Problem be converted to Node-Weighted Steiner Tree Problem?
Sounds right. Denote by $G$ the original graph, and by $G'$ the graph after introducing the new vertices.

The first direction is trivial - every Steiner tree in $G$ corresponds to a Steiner tree in $G'$, with the same weight (by splitting each edge with its new vertex).

The second direction is less trivial. Still, consider a minimal Steiner tree $T$ in $G'$. We can assume w.l.o.g that $T$ does not contain any leaf that corresponds to a new edge-vertex. Indeed, this vertex is not in $S$, so if it is a leaf, it can be removed from the tree, keeping it a Steiner tree with minimal weight (unless you allow negative weights, which I assume you don't).

Now that we have a Steiner tree without edge-vertex leafs, we can convert it to a Steiner tree in $G$: root the tree by some vertex that is not an edge-vertex, then by the construction, all the edge-vertices have a single child in this tree. So you can replace each pair of edges $(u,X),(X,v)$ by $(u,v)$ in the original graph, obtaining a Steiner tree with the same weight.

I don't see any bugs here, but maybe I missed something.