fix: passing ResultError to Error thru operator<<

It is typical to pass error to callers like following;

    if (!result) {
      return Error() << result.error();
    }

To transfer errno(or ResultError#code()), Error defines a specialization
operator<<(const ResultError&).

This change fixes so that ResultError is properly handled

Bug: 132145659
Test: atest libbase_test
Change-Id: Ib35457da2d4b923d8e652c54ac510a75546cf918