To summarize the argument in more general terms: the class of algorithms in P is those where you can derive some global property in terms of increment local decisions. For example, building the shortest path between two nodes in a graph can be done by always picking the closest node. By contrast, NP-hard algorithms have the problem that incremental local decisions don't let that happen: you might need to recolor an entire optimally-colored subgraph to admit a new node.
Furthermore, we know from research that there tends to be a very sharp transition from P to NP-hard in terms of transformation, where relaxing a single condition goes straight from P to NP-hard without any intermediate "we don't know where there is" space. This tends to suggest that it's not really so much a question of problems being P or NP, but rather of instances being intrinsically easy to hard.
The question of P=NP then is really about whether these hard instances are really exponentially hard, or can we embed them into a simpler instance using some combinatorial construct that merely makes it look exponentially hard. So it's still possible for P=NP, but the practical question of "will we get an efficient, guaranteed for all instances, solution to these problems?" is answered in the negative. When you do the combinatorial embeddings to make the polynomial time algorithms work, you end up getting constants that look like 3^2^2^4, and there's no way to shrink those constants to practical ones.