Does an expression with undefined behaviour that is never actually executed make a program erroneous?

If a side effect on a scalar object is unsequenced relative to etc

Side effects are changes in the state of the execution environment (1.9/12). A change is a change, not an expression that, if evaluated, would potentially produce a change. If there is no change, there is no side effect. If there is no side effect, then no side effect is unsequenced relative to anything else.

This does not mean that any code which is never executed is UB-free (though I'm pretty sure most of it is). Each occurrence of UB in the standard needs to be examined separately. (The stricken-out text is probably overly cautious; see below).

The standard also says that

A conforming implementation executing a well-formed program shall produce the same observable behavior as one of the possible executions of the corresponding instance of the abstract machine with the same program and the same input. However, if any such execution contains an undefined operation, this International Standard places no requirement on the implementation executing that program with that input (not even with regard to operations preceding the first undefined operation).

(emphasis mine)

This, as far as I can tell, is the only normative reference that says what the phrase "undefined behavior" means: an undefined operation in a program execution. No execution, no UB.

No. Example:

struct T {
    void f() { }
};
int main() {
    T *t = nullptr;
    if (t) {
        t->f(); // UB if t == nullptr but since the code tested against that
    }
}

Deciding whether a program will perform an integer division by 0 (which is UB) is in general equivalent the halting problem. There is no way a compiler can determine that, in general. And so the mere presence of possible UB can not logically affect the rest of the program: a requirement to that effect in the standard, would require each compiler vendor to provide a halting problem solver in the compiler.

Even simpler, the following program has UB only if the user inputs 0:

#include <iostream>
using namespace std;

auto main() -> int
{
    int x;
    if( cin >> x ) cout << 100/x << endl;
}

It would be absurd to maintain that this program in itself has UB.

Once the undefined behavior occurs, however, then anything can happen: the further execution of code in the program is then compromised (e.g. the stack might have been fouled up).