Can the isomorphic graph problem be solved in deterministic polynomial time?
Your approach sounds intuitively appealing on first glance, but it doesn't work.  It's not enough that each edge in $G$ has a matching edge in $H$, when taken one at a time.  There has to be a way to provide a single matching (isomorphism) that matches every edge of $G$ to a (different) edge of $H$, in a way that preserves the connectivity.

Examples:


Suppose you discover that it is possible to match edge $(a,b)$ from $G$ with edge $(u,v)$ from $H$; and you also discover that it is possible to match edge $(c,d)$ from $G$ with $(u,v)$ from $H$.  This doesn't mean there will necessarily be a way to stitch this into a complete isomorphism that describes how all vertices and edges are matched.  For instance, trying to simultaneously match both $(a,b)$ and $(c,d)$ to the same edge $(u,v)$ does not correspond to any legal graph isomorphism (different vertices have to map to different vertices, and different edges to different edges).
Suppose you discover that it is possible to match edge $(a,b)$ from $G$ with edge $(u,v)$ from $H$; and you also discover that it is possible to match edge $(b,c)$ from $G$ with $(w,x)$ from $H$, where $v \ne w$.  This doesn't mean there will necessarily be a way to stitch this into a complete isomorphism that describes how all vertices and edges are matched.  For instance, trying to simultaneously match $(a,b)$ to $(u,v)$ and $(b,c)$ to $(w,x)$ doesn't work, since you need $b$ to be consistently matched to the same vertex in both (e.g., either to $v$ both times, or to $w$ both times).


For these reasons, your approach will indeed run in polynomial time but will not compute the correct result: there will be some graphs $G,H$ where your algorithm outputs "isomorphic" but where the correct answer is "not isomorphic".



If we try to fix this problem, the naive approach is to try to find all ways to match the set of $n$ vertices in $G$ to a re-ordering of the set of vertices in $H$ (i.e., a matching that simultaneously describes what every vertex in $G$ corresponds to, in $H$).  However, there are $n!$ such candidates to try, and $n!$ is exponential in $n$, so this leads to an exponential-time algorithm.  It's a famous open problem whether the graph isomorphism problem is in P or not.  See https://en.wikipedia.org/wiki/Graph_isomorphism.