I will prove this using induction, with the assumption that the commutative and associative property for at most $3$ numbers have been proved before.
Base case: If $n = 1$, then $\sum_{i=1}^n a_i = a_1$. Moreover, there is only one possible permutation $\sigma$: $\sigma(1) = 1$. Therefore, $\sum_{i=1}^n a_{\sigma(i)} = a_{\sigma(1)} = a_1$ as well. Hence, we have the required statement.
If $n = 2$, then $\sum_{i=1}^n a_i = a_1 + a_2$. There are two possible options on what $\sigma(1)$ could be.
If $\sigma(1) = 1$ then $\sigma(2) = 2$. In this case, $$ \sum_{i=1}^n a_{\sigma(i)}= a_{\sigma(1)} + a_{\sigma(2)} = a_1 + a_2. $$ If $\sigma(1) = 2$ then $\sigma(2) = 1$. Similarly, we have $$ \sum_{i=1}^n a_{\sigma(i)}= a_{\sigma(1)} + a_{\sigma(2)} = a_2 + a_1. $$ Combining these facts with the commutative property, we can conclude that $\sum_{i=1}^n a_{\sigma(i)} = \sum_{i=1}^n a_{i}$ is true when $n = 2$.
Induction step: Assume that the statement is true for every natural number up to $k$. Let's investigate the case where $n = k + 1$.
By definition, we have: $$ \sum_{i=1}^{k+1}a_{\sigma(i)} = \sum_{i=1}^{k}a_{\sigma(i)} + a_{\sigma(k+1)} \text{ and } \sum_{i=1}^{k+1}a_{i} = \sum_{i=1}^{k}a_{i} + a_{k+1}. $$
If $\sigma(k+1) = k+1$, then $\sigma$ is also a permutation on $I_k$, not just $I_{k+1}$. Using the induction hypothesis, $\sum_{i=1}^{k}a_{\sigma(i)} = \sum_{i=1}^{k}a_{i}$ and hence $\sum_{i=1}^{k+1}a_{\sigma(i)} = \sum_{i=1}^{k+1}a_{i}.$
In the other case where $\sigma(m) = k+1$ and $m\neq k + 1$, create a $k$-length sequence $(b_1, b_2, \cdots, b_k)$ defined as $b_i = a_{\sigma(i)}$.
Let $s_w$ be a permutation on $I_k$ such that $$ \begin{cases} s_w(i) = i \text{ if } i\notin \{m; k\} \\ s_w(m) = k \\ s_w(k) = m \end{cases}. $$
Using the inductive hypothesis, we have: $$ \sum_{i=1}^{k+1}a_{\sigma(i)} = \sum_{i=1}^{k}b_{i} + a_{\sigma(k+1)} = \sum_{i=1}^{k}b_{s_w(i)} + a_{\sigma(k+1)}. $$ This implies $$ \sum_{i=1}^{k+1}a_{\sigma(i)} = \sum_{i=1}^{k}a_{\sigma\left(s_w(i)\right)} + a_{\sigma(k+1)} = \sum_{i=1}^{k-1}a_{\sigma\left(s_w(i)\right)} + a_{\sigma\left(s_w(k)\right)} + a_{\sigma(k+1)}=\left(\sum_{i=1}^{k-1}a_{\sigma\left(s_w(i)\right)} + a_{\sigma(k+1)}\right) + a_{k+1}.$$
Let $\tilde{\sigma}: I_k \to I_{k+1}$ such that $\begin{cases} \tilde{\sigma}(i) = \sigma\left(s_w(i)\right) \text{ if } i < k \\ \tilde{\sigma}(k) = \sigma(k+1) \end{cases}$. However, $\tilde{\sigma}(i) \neq k + 1$ as $\sigma\left(s_w(i)\right) = k + 1$ implies $i = k$ and $\sigma(k+1) \neq k +1$. Therefore, $\tilde{\sigma}$ is equivalent to a mapping from $I_k$ to $I_k$.
As $s_w$ and $\sigma$ are 2 permutations and $s_w(i) \neq k+1$ for every $i$, $\tilde{\sigma}$ is injective. Also, $\tilde{\sigma}$ is subjective. Let $y \in I_{k}$. There exists $x \in I_{k+1}$ such that $\sigma(x) = y$. If $x = k+1$ then $y = \tilde{\sigma}(k)$. Otherwise, there exists $p$ in $I_k$ such that $\sigma(s_w(p)) = y$. With a notice that $p \neq k$, $p \in I_{k-1}$ and hence $\tilde{\sigma}(p) = y$.
In short, $\tilde{\sigma}$ is a permutation on $I_k$. Therefore, $$ \sum_{i=1}^{k+1}a_{\sigma(i)} = \left(\sum_{i=1}^{k-1}a_{\sigma\left(s_w(i)\right)} + a_{\sigma(k+1)}\right) + a_{k+1} = \sum_{i=1}^{k}a_{\tilde{\sigma}(i)} + a_{k+1} = \sum_{i=1}^{k}a_{i} + a_{k+1} = \sum_{i=1}^{k+1}a_{i}. $$
The induction step is complete and we have what need to be proven.
