Does a program with std::map<T*, U> have well-defined behaviour?

Yes, because std::map default comparison operator is std::less, which, unlike the standard comparison operator, is completely defined for pointer types.

[comparisons#2]

For templates less, greater, less_­equal, and greater_­equal, the specializations for any pointer type yield a result consistent with the implementation-defined strict total order over pointers ([defns.order.ptr]).

The implementation-defined strict total order over pointers is defined in [defns.order.ptr] as:

implementation-defined strict total ordering over all pointer values such that the ordering is consistent with the partial order imposed by the builtin operators <, >, <=, >=, and <=>

std::less (default comparer of std::map) has special treatment about pointer allowing that:

A specialization of std::less for any pointer type yields a strict total order, even if the built-in operator< does not.

And about

can we say that a map with pointer keys is a dangerous proposition?

So it is fine in general.

Additional precaution should be taken with const char* key:

We compare pointers and not string content (mostly beginner confusions).

C-string literals with same content have no guaranty to be equals:

"literal" == "literal"; // Not guaranteed
"literal" < "literal"; // false .. or true

std::map use std::less that have a specialization for pointer type :

A specialization of std::less for any pointer type yields a strict total order, even if the built-in operator< does not. The strict total order is consistent among specializations of std::less, std::greater, std::less_equal, and std::greater_equal for that pointer type, and is also consistent with the partial order imposed by the corresponding built-in operators (<, >, <= and >=).

For a more specific description i leave you 2 links:

std::less first link

std::less second link