Let the numbers be $a, a+b, a+2b, a+3b$ and $a+4b$. If we assume $b=0$, then $a<11$ based on the wrong product. Even where numbers are concatenated, the concatenated number's mod 9 value is equivalent to the sum of the initial numbers, so the the sum being wrong doesn't make a difference in this aspect. The wrong sum can be written as $9k+3$, meaning the average ($a+2b$) must be written as $9k+6$. Since the digit of ones of $a+4b$ is the last ever digit in the multiplication, it must be odd, making also $a$ and $a+2b$ odd. $a$ can only be an odd digit. According to the last sentence in the question, $a+b$ must have multiple digits. 71157 must be the sum of 3 terms, so at least one of them has 5 digits. Since $a+b > (a+4b)/4$, $a+4b$ can have one digit more than $a+b$ at most. If $a+b$ has $d$ digits, $2d<6$, $d<3$, so $d$ can only be 2 and $a+b$ is a two-digit number whose digit of tens is 1. Even $a+4b$ has 2 digits. A 5-digit number can only be formed by concatenating three consecutive numbers, with the first being a single-digit number ($a$ that is).
The remaining terms in the addition are the last two numbers, whose sum is less than 199, making the $a$ in the 5-digit number 7.
$a+b$ starts with 1, which becomes the digit of thousands of the 5-digit number. That means $a+3b+a+4b = 2a+7b < 158$
That 5-digit number ends with $a+2b$, an odd 2-digit number that can be written as $9k+6$ whose digit of ones is at most 3. May be 15 or 33, but the latter means $a+b=20$, so $a+2b=15$ and $b=4$.
The numbers are 7, 11, 15, 19 and 23. The addition is $71115+19+23=71157$ and the multiplication is $7*11*151923=11698071$.