"is not required" == undefined behavior?

valid C++ code

Yes.

Nowhere does the standard say that this is UB or ill-formed, and neither is this case lacking a rule describing the behaviour because the quoted 4.3 applies.

and the output is either 11 or 00

I'm not sure that 10 or 01 are technically guaranteed to not be output ¹.

Given that neither pointer is required to compare greater than the other, the result of the comparison can be either true of false. There appears to not be an explicit requirement for the result to be the same for each invocation on same operands in this case.

¹ But I consider this unlikely in practice. I also think that leaving such possibility open is not intentional. Rather, the intention is to allow for deterministic, but not necessarily total order.

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P.S.

auto comp = std::less<>;

std::cout << comp(&x, &y);
std::cout << comp(&x, &y);

would be guaranteed to be either 11 or 00 because std::less (like its friends) is guaranteed to impose a strict total order for pointers.

For the provided code, this case applies:

(4.3) Otherwise, neither pointer is required to compare greater than the other.

There is no mention of UB, and so a strict reading of "neither is required" suggests that the result of the comparison could be different every time it's evaluated.

This means the program could validly output any of the following results:

00
01
10
11

The wording has changed in various editions of the C++ standard, and in the recent draft cited in the question. (See my comments on the question for the gory details.)

C++11 says:

Other pointer comparisons are unspecified.

C++17 says:

Otherwise, neither pointer compares greater than the other.

The latest draft, cited in the question, says:

Otherwise, neither pointer is required to compare greater than the other.

That change was made in response to an issue saying ""compares greater" term is needlessly confusing".

If you look at the surrounding context in the draft standard, it's clear that in the remaining cases the result is unspecified. Quoting from [expr.rel] (text in italics is my summary):

The result of comparing unequal pointers to objects is defined in terms of a partial order consistent with the following rules:

• [pointers to elements of the same array]

• [pointers to members of the same object]

• [remaining cases] Otherwise, neither pointer is required to compare greater than the other.

If two operands p and q compare equal, p<=q and p>=q both yield true and p<q and p>q both yield false. Otherwise, if a pointer p compares greater than a pointer q, p>=q, p>q, q<=p, and q<p all yield true and p<=q, p<q, q>=p, and q>p all yield false. Otherwise, the result of each of the operators is unspecified.

So the result of the < operator in such cases is unspecified, but it does not have undefined behavior. It can be either true or false, but I don't believe it's required to be consistent. The program's output could be any of 00, 01, 10, or 11.