For each integer $j$ relatively prime to $n$, let $r_j$ be the remainder when $aj$ is divided by $n$. Then we have
$$\begin{aligned} a^{\varphi(n)} &= \prod_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{aj}{j} = \prod_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{1}{j} \left( \left\lfloor \frac{aj}{n} \right\rfloor n + r_j \right) \\ &= \prod_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{r_j}{j} \left( 1 + \frac{n}{r_j} \left\lfloor \frac{aj}{n} \right\rfloor \right) = \prod_{\substack{j=1 \\ (j,n)=1}}^{n} \left( 1 + \frac{n}{r_j} \left\lfloor \frac{aj}{n} \right\rfloor \right) \end{aligned}$$
since $\{r_1, r_2, \dots, r_{\varphi(n)}\}$ is a reduced set of residues modulo $n$. Expanding the product and taking modulo $n^2$, we obtain
$$a^{\varphi(n)} \equiv 1 + n \sum_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{1}{r_j} \left\lfloor \frac{aj}{n} \right\rfloor \equiv 1 + n \sum_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{1}{aj} \left\lfloor \frac{aj}{n} \right\rfloor \pmod{n^2}.$$
This implies
$$\frac{a^{\varphi(n)} - 1}{n} \equiv \sum_{\substack{j=1 \\ (j,n)=1}}^{n} \frac{1}{aj} \left\lfloor \frac{aj}{n} \right\rfloor \pmod{n}.$$