As @coffeemath noted, your answer is actually the complement of the original expression. So this method is not correct.
I think the easiest way is to use the identities $(x + y)' = x'y'$ and $(xy)' = x' + y'$ (known as De Morgan's laws) and $x'' = x$. We have $$AB' + A'B = ((AB')'(A'B)')' = ((A'+B)(A+B'))'=$$
$$=(A'A + A'B' + AB + BB')' = (AB + A'B')' = (AB)'(A'B')' = (A'+B')(A+B).$$
Hence the original expression equals $$ (AB'+A'B)C = (A'+B')(A+B)C.$$
Alternatively, if you're familiar with how to get a SOP or a POS from the truth table, then you can just construct the truth table for the function $f(A, B)$ realized by the expression $AB' + A'B$ and construct a POS directly from it. In our case the truth table looks like this:
A & B & f(A, B) \\ \hline
0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\ \hline
To construct the POS representation of $f(A, B)$ you should look at pairs $(a, b)$ with $f(a, b) = 0$. With every such pair you associate a sum $A^\sigma + B^\tau$, where $A^\sigma = A$ if $a = 0$ and $A^\sigma = A'$ if $a = 1$, and same for $B^\tau$. After that you just construct the product of all these sums. In our case we have two pairs $(0, 0)$ and $(1, 1)$. Corresponding sums are $A + B$ and $A' + B'$. Finally, we have the POS $(A'+B')(A+B)$.