$(1)$Using the Dirichlet convolution theorem, $$\sum_{n|d}\mu(d)\frac{n}{d}=\sum_{ab=n}\mu(a)b$$ Let $n=p_1\cdot p_2\cdots p_n\cdot k$, where $k$'s prime factorisation contains $p_k$ only if $ k\in\{1,2,...,n\}$. Note that, for nonzero $\mu(a)$, $k|b$, else $a$ wouldn't be square-free. $(2)$ Thus there is a bijection between all $a$ such that $k|b$ and all subsets ($S$) of $\{p_1,p_2,...,p_n\}$. If $|S|$ is odd, $\mu(a)=-1$, if even, $\mu(a)=1$. Hence the original sum is equal to, $$\displaystyle (-1)^nk\left(1+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b \end{matrix} }p_ap_b+\sum_{\begin{matrix} 1\le a,b,c,d \le n\\ a<b<c<d \end{matrix}}p_ap_bp_cp_d+\cdots\right)$$ $$\displaystyle (-1)^{n+1}k\left(\sum_{ 1\le a \le n}p_a+\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c \end{matrix} }p_ap_bp_c+\cdots\right)$$ $$=\displaystyle (-1)^nk\left(1-\sum_{ 1\le a \le n}p_a+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b \end{matrix} }p_ap_b-\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c \end{matrix} }p_ap_bp_c+\sum_{\begin{matrix} 1\le a,b,c,d \le n\\ a<b<c<d \end{matrix}}p_ap_bp_cp_d-\cdots\right)$$
$$=\displaystyle (-1)^nn\left(\frac{1}{p_1\cdots p_n}-\sum_{ 1\le a \le n}\frac{p_a}{p_1\cdots p_n}+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b \end{matrix} }\frac{p_ap_b}{p_1\cdots p_n}-\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c \end{matrix} }\frac{p_ap_bp_c}{p_1\cdots p_n}+\cdots\right)$$
This looks quite a lot like the inclusion-exclusion principle might work well here to show that $\sum_{ab=n}\mu(a)b=\varphi(n)$, but I'm not well versed enough in the principle to do so. Could anyone use the principle from here to prove the theorem? Algebraically, it's possible to prove the initial theorem algebraically using
$$\displaystyle(-1)^n\left(\frac{1}{p_1\cdots p_n}-\sum_{ 1\le a \le n}\frac{p_a}{p_1\cdots p_n}+\sum_{\begin{matrix} 1\le a,b \le n\\ a<b \end{matrix} }\frac{p_ap_b}{p_1\cdots p_n}-\sum_{\begin{matrix} 1\le a,b,c \le n\\ a<b<c \end{matrix} }\frac{p_ap_bp_c}{p_1\cdots p_n}+\cdots\right)=\prod_{k=1}^n\left(1-\frac{1}{p_k}\right), $$
but it would be nice to have an inclusion-exclusion (I think this is combinatorial) argument.
Let, $n=p_1^{e_1}p_2^{e_2}...p_k^{e_k}$. Now by definition,$$\varphi(n)=\#\{m:m\le n, (m,n)=1\}.$$Now if, $m<n$ such that $(m,n)=1$ then, $p_i\nmid m$ for any $i\le k.$ Therefore, number of such $m$ would be, by inclusion-exclusion principle,$$n-\sum_iA_{p_i}+\sum_{i<j}A_{p_ip_j}-\sum_{i<j<k}A_{p_ip_jp_k}...$$ where $A_d=\#\{m:m\le n,d|m\}=\frac nd$.So,$$\varphi(n)=n-\sum_i\frac n{p_i}+\sum_{i<j}\frac n{p_ip_j}...=\sum_{d|n}\mu(d)\frac n d.$$