The sets $\{ 1, 2, 3 \} $ and $\{ 0, 1, 2 \} $ are different sets. They have different elements.
Back to the question. For any set $X$ of ordinals there is an ordinal that is greater than or equal to each and every ordinal in $X$. Since the ordinals are well-ordered there is a least ordinal that is greater than or equal to each and every ordinal in $X$. This ordinal is the supremum of $X$. If $X$ happens to have a largest element then that largest element is the supremum of $X$. (Compare this with the real numbers.)
For your problem there are two cases: either $A$ has a largest element or $A$ does not have a largest element.
Suppose the first case holds. Let $\xi^*$ be the largest element of $A$. The element $\xi^* + 1$ is larger than every element in $A$ and is the supremum of the elements of $\{ \xi + 1 \colon \xi \in A \} $. Any ordinal that is strictly smaller than $\xi^*$ will be no larger than at least one element of $A$, namely $\xi ^*$. In this case we have the desired result.
Suppose the second case holds. In this case $\sup A$ is the least ordinal greater than every ordinal in $A$. To see this choose $\xi_{0} \in A$. By our assumption that the second case holds there is a $\xi_{1} \in A$ satisfying $\xi_{0} < \xi_{1} \leq \sup A$. Since $\xi_{0}$ was an arbitrarily chosen element of $A$ we get $\sup A$ is the least ordinal greater than every element of $A$. Now we show that $\{ \xi + 1 \colon \xi \in A \} = \sup A$. Select $\xi_{0} \in A$. Not that $\xi_{0} < \xi_{0} + 1 \leq \xi_{1}$, where $\xi_{1}$ is some ordinal in $A$ that is greater than $\xi_{0}$. Since for every element $\alpha \in B = \{ \xi + 1 \colon \xi \in A \} $ there is another element of $A$ which is greater than $\alpha$ the supremum of $A$ is no smaller than the supremum of $B$. But then the two supremum are equal.