I'm quoting Boolos, Burgess, and Jeffrey's Computability and Logic
Consider the language $L^*$ of arithmetic and three different interpretations of it: first, the standard interpretation $\mathcal{N}^*$; second, the alternative interpretation $\mathcal{Q}$ we considered earlier, with domain the nonnegative rational numbers; third, the similar alternative interpretation $\mathcal{R}$ with domain the nonnegative real numbers. Now in fact the substitutional approach gives the intuitively correct results for $\mathcal{N}^*$ in all cases. Not so, however, for the other two interpretations. For, all closed terms in the language have the same denotation in all three interpretations, and from this it follows that all closed terms denote natural numbers. And from this it follows that $t + t = 1$ is false for all closed terms $t$, since there is no natural number that, added to itself, yields one.
The "substitutional approach" mentioned states
$$ M\models \forall xF(x) \iff \text{ for all closed terms } t, M\models F(t)$$ $$ M\models \exists xF(x) \iff \text{ there exists a closed term } t, M\models F(t)$$
I don't see how all closed terms in the language have the same denotation in all three interpretations. Clearly if this is true, then it follows from this that the closed terms denote the natural numbers (since there is nothing else in $\mathcal{N}^*$).
There should be a variety of ways that you can see that any interpretation of a closed term in $\mathcal{Q}$ or $\mathcal{R}$ (assuming they are what I imagine them to be) is the same value you get by interpreting it in $\mathcal{N}^*$ and then composing with the relevant inclusion $\mathbb{N} \subseteq \mathbb{Q}_{\geq 0} \subseteq \mathbb{R}_{\geq 0}$
But to argue from scratch, you can proceed by structural induction. I assume that the closed terms in the language you refer to are those term that can be produced from the constants $0$ and $1$ by applying the binary operation symbols $+$ and $\cdot$.
The base case of the induction is on the constants: $0$ and $1$ are both natural numbers when interpreted by $\mathcal{Q}$ or $\mathcal{R}$. The inductive step is, assuming the interpretations of $\alpha$ and $\beta$ are both natural numbers, we know that the interpretations of $\alpha + \beta$ and $\alpha \cdot \beta$ are natural numbers too, since both $\mathcal{Q}$ and $\mathcal{R}$ both have the property that the sum and product of natural numbers are both natural numbers.