Are there "interesting" theorems in Peano arithmetic, that only use the addition operation?

For fixed relatively prime integers $m$ and $n$, the fact that a number divisible by both of them is divisible by their product is expressible in Presburger arithmetic. The case $m=2$, $n=3$:

$$\forall x[(\exists y)(y+y=x)\land(\exists y)(y+y+y=x)\to(\exists y)(y+y+y+y+y+y=x)]$$

Presburger Arithmetic can express $x\equiv y\pmod n$ for fixed $n$, so we can state instances of the Chinese Remainder Theorem for fixed moduli, such as

For all $a$ and $b$ there is an $x$ such that $x\equiv a\pmod{7}$ and $x\equiv b\pmod{8}$.

Since this claim can be expressed and it's true in $\mathbb N$, Presburger Arithmetic, being complete, must be able to prove it.

The associativity of addition is one candidate. Recall from the definition of addition that \begin{align} a+0 &=a & \text{A1} \\ a+S(b) &=S(a+b) & \text{A2} \end{align}

Theorem: $(a+b)+c = a+(b+c)$

Proof: Let $\phi(c)$ denote the proposition that $(a+b)+c = a+(b+c)$. Proceed by induction on $c$.

Base case: Let $c=0$. Then \begin{align} (a+b)+c &= (a+b)+0 && \text{A1} \\ &= a+b \\ &= a+(b+0) && \text{A1} \\ &= a+(b+c) \end{align}

Therefore $\phi(0)$.

Inductive step: Suppose $(a+b)+c = a+(b+c)$. Then \begin{align} (a+b)+S(c) &= S((a+b)+c) && \text{A2} \\ &= S(a+(b+c)) && \text{inductive hypothesis} \\ &= a+S(b+c) && \text{A2} \\ &= a+(b+S(c)) && \text{A2} \end{align}

Therefore $\phi(c) \rightarrow \phi(c+1)$ for all $c \in \mathbb{N}$.

By the axiom schema of induction, $\phi(c)$ for all $c \in \mathbb{N}$. This proves that $(\mathbb{N}, +)$ is a semigroup. Together with the proof that 0 is a left identity (in addition to being a right identity as stated in A1), this proves that $(\mathbb{N}, +)$ is a monoid.