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It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example: $S = -\mathbb{N}$, which is clearly not an additive basis, but $S^2 = \mathbb{N}^2$ obviously is, by Lagrange's 4 squares theorem.

I believe this answers your question in all 4 possible cases!

It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Edit: this only works if $S$ is a finite set, otherwise my proof does not hold

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example: $S = -\mathbb{N}$, which is clearly not an additive basis, but $S^2 = \mathbb{N}^2$ obviously is, by Lagrange's 4 squares theorem.

I believe this answers your question in all 4 possible cases!

It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example $S = \{-1\}$, which is clearly not an additive basis, but $S^2 = \{1\}$ obviously is.

I believe this answers your question in all 4 possible cases!

It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example: $S = -\mathbb{N}$, which is clearly not an additive basis, but $S^2 = \mathbb{N}^2$ obviously is, by Lagrange's 4 squares theorem.

I believe this answers your question in all 4 possible cases!

It would be good to get clarification as to whether $S$ must be positive integers, or are they allowed to be negative too? If they must be positive, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work. I believe I have a proof that $S \cup S^2$ is an additive basis if and only if $S$ is. The if direction is clear, we will show $S \cup S^2$ additive basis implies $S$ additive basis.

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

It would be good to get clarification as to whether $S$ must be positive integers, and whether you need the summands to be distinct? If they must be positive and distinct, $S = 3\mathbb{N} \cup \{2\}$ works as an example for a non asymptotic additive basis where $S \cup S^2$ is an asymptotic additive basis.

$S$ is not an asymptotic additive basis, as any sum of elements in $S$ is either a multiple of 3 or 2 more than a multiple of 3. But, since $S^2$ also contains 4, we can now write all integers other than 1 as a sum of at most 3 elements from $S \cup S^2$, which is therefore an asymptotic additive basis.

If you are allowed negative elements, then the previous example with $\mathbb{N}$ replaced with $\mathbb{Z}$ works for the stronger statement of additive basis (actually, you only need to add -1 to the set $S$ above for this).

If you are allowed to 're use' elements, this example doesn't work, and I have a proof no example will work if $S$ must be positive integers. We will show $S \cup S^2$ additive basis implies $S$ additive basis.

Assume $S \cup S^2$ is an additive basis, let $n \in \mathbb{N}$. Then there exist $x_1, \ldots, x_k, y_1, \ldots, y_m \in S$ such that $$ n = x_1 + \cdots + x_k + y_1^2 + \cdots + y_m^2 $$ But then we can write each $y_i^2$ as $y_i + \cdots + y_i$, where there are $y_i$ summands (that is, add $y_i$ to itself $y_i$ times).

Hence, any $n \in \mathbb{N}$ can be written as a sum of $x_i$ and $y_i$ which are in $S$, and so $S$ is an additive basis.

Finally, if $S$ does not need to be positive integers, but you can re use elements of $S$, then there is a trivial example $S = \{-1\}$, which is clearly not an additive basis, but $S^2 = \{1\}$ obviously is.

I believe this answers your question in all 4 possible cases!