Why is "++i++" invalid while (++i)++ is valid?

i and ++i are both lvalues, but i++ is an rvalue.

++(i++) cannot be valid, as the prefix ++ is being applied to i++, which is an rvalue. But (++i)++ is fine because ++i is an lvalue.

Note that in C, the situation is different; i++ and ++i are both rvalues. (This is an example of why people should stop assuming that C and C++ have the same rules. People insert these assumptions into their questions, which must then be refuted.)

so the code is parsed as int b = ++(i++); and i is an rvalue.

No. i is not an rvalue. i is an lvalue. i++ is an rvalue (prvalue to be specific).

This declaration

int b = ++i++;

is equivalent to

int b = ++( i++ );

The postfix increment operator returns the value of the operand before increment.

From the C++ 17 Standard (8.2.6 Increment and decrement)

1 The value of a postfix ++ expression is the value of its operand...The result is a prvalue.

While the unary increment operator returns lvalue after its increment. So this declaration

int b = (++i)++;

is valid. You could for example write

int b = (++++++++i)++;

From the C++ 17 Standard (8.3.2 Increment and decrement)

1 The operand of prefix ++ is modified by adding 1. The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type other than cv bool, or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field....

Pay attention that in C the both operators return a value instead of lvalue. So in C this declaration

int b = (++i)++;

is invalid.