Let me give a construction that leads, for any fixed $N$, to a graph which has $N$ pairwise non-isomorphic edges and removal of any one of them results in the same graph. This is not a complete answer but I think it is of some interest.
Let us denote by $G$ a graph whose vertices are labeled with elements of a finite set $V$. We will assume that $\pi=\{p,q\}$ and $\rho=\{r,s\}$ is a non-isomorphic pair of edges in $G$. I will also assume that the graph $G'=G\setminus\pi$ (obtained from $G$ by removing $\pi$) is isomorphic to $G\setminus\rho$. For example, the 6-vertex graph from the initial question can be taken as $G$.
Let me construct the graph $\Gamma=\Gamma(G,\pi)$ as follows. The vertex set of $\Gamma$ will be the union of $V$ and $W\times V$; we denote the vertices from $W\times V$ as $v_{ec}$ for $e\in W$ and $c\in V$. (Here, $W$ stands for the set of all subsets of $V$ which have cardinality $2$.) The edges of $\Gamma$ are the following.
(1) there is an edge between $i$ and $v_{ec}$ if and only if $i\in e$;
(2) assuming $e\in G$, there is an edge between $v_{ex}$ and $v_{ey}$ if and only if $\{x,y\}\in G$;
(3) assuming $e\notin G$, there is an edge between $v_{ex}$ and $v_{ey}$ if and only if $\{x,y\}\in G'$;
(4) $\Gamma$ has only those edges which are specified in (1)-(3).
Speaking informally, we draw a copy of $G$ instead of every edge $\{i,j\}$. If $\{i,j\}$ is not an edge, we draw a copy of $G'$. In both cases, we draw the edges connecting all the new vertices with both $i$ and $j$. (We do this over all $\{i,j\}\in W$.)
Now, consider the edges $\{v_{\pi p},v_{\pi q}\}$, $\{v_{\pi r},v_{\pi s}\}$, $\{v_{\rho p},v_{\rho q}\}$, $\{v_{\rho r},v_{\rho s}\}$. From an informal description it can be seen that graphs obtained by removal of any one of them are isomorphic. To show that the edges provided are pairwise non-isomorphic, assume by contradiction that an automorphism $\psi$ of $\Gamma$ swaps some pair of them. Then,
(1) a vertex $v_{ec}$ can not be an image $\psi(i)$ of $i$ under $\psi$ because they have different degrees;
(2) then, denoting $e=\{i,j\}$ and $\psi(e)=\{\psi(i),\psi(j)\}$, we conclude $\psi(v_{ec})=v_{\psi(e)c'}$ for some $c'$;
(3) the subgraphs generated by all $v_{ec}$ and by all $v_{e'c}$ (in both cases, $c$ runs over $1,\dots,n$) are isomorphic iff either $e,e'\in G$ or $e,e'\notin G$, therefore $\{\psi(i),\psi(j)\}\in G$ iff $\{i,j\}\in G$;
(4) this shows that $\psi'$, the restriction of $\psi$ to $\{1,\ldots,n\}$, is an automorphism of $G$;
(5) by the definition of $G$, $\psi'$ can not swap $\pi$ and $\rho$, so that $\psi(\pi)=\pi$ and $\psi(\rho)=\rho$;
(6) then, the restriction of $\psi$ to $v_{\pi c}$ is an automorphism, which cannot swap $\{v_{\pi p},v_{\pi q}\}$ and $\{v_{\pi r},v_{\pi s}\}$ again by definition of $G$;
(7) similarly to the previous step, $\psi$ cannot swap $\{v_{\rho p},v_{\rho q}\}$ and $\{v_{\rho r},v_{\rho s}\}$;
(8) finally, $\psi$ can not swap $v_{\rho a}$ and $v_{\pi b}$ because of step (5).
Let me finally note that we can apply the $\Gamma$ construction to $\Gamma_0=\Gamma(G,\pi)$ itself. As far as I can see, the graph $\Gamma_1=\Gamma(\Gamma_0,\{v_{\rho p},v_{\rho q}\})$ has $16$ pairwise non-isomorphic edges and removal of any one of them results in the same graph. Similarly constructed $\Gamma_2$ will have $256$ edges with this property, and so on.
Best Answer
I think one can push through the probabilistic arguments of Tim Gowers and Fedor Petrov in the general case, as follows.
Let $c$ be a constant such that the number of red edges in $G[S]$ is at most $c|S|$ for every $S \subseteq V(G)$. One can order the vertices of $G$: $v_1, v_2, \ldots, v_n$, so that every vertex has at most $2c$ neighbors with lower indices. (Define the ordering starting with the highest index. If $v_n, \ldots,v_{i+1}$ are defined, set $v_i$ to be the vertex with the smallest degree in the subgraph induced by the vertices which are not yet indexed. This is a standard trick.)
Now we define a random subset $S$ of $V(G)$ recursively: if $S \cap$ {$v_1, \ldots, v_i$} is chosen put $v_{i+1}$ in $S$ with probability $1/2$ if it is not joined by a red edge to any of the vertices already in $S$, otherwise don't put it in $S$. Then $S$ is red-free and, just as in Fedor's answer, we can see that the probability that a pair of vertices $u$ and $v$ joined by a blue edge both lie in $S$ is at least $2^{-4c-2}$. Therefore the number of blue edges is at most
$2^{4c+2}c' \mathbf{E}[|S|] \leq 2^{4c+1}c'|V(G)|,$
where $c'$ is the constant implicitly present in the condition on the density of the blue edges.