The problem was in how to write down the sum $\sum_{\{\vec{S}\}}$. Since this sum is over all the possible vectors $\vec{S}=(S_1,...,S_N)$, (where $S_i\in\{-1,1\}$), we can rewrite this sum like
$$\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}$$
i.e.
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}\prod_{i=1}^{N}e^{\beta HS_{i}} \qquad(1)$$
Clearly this sum has $2^N$ elements of the type $\prod_{i=1}^{N}e^{\beta HS_{i}}$. Now, since
$$\prod_{i=1}^{N}e^{\beta HS_{i}}=e^{\beta HS_{1}}\cdots e^{\beta HS_{N}},$$
then (1) turns
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\sum_{S_{1}\in\{-1,1\}}\cdots\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{1}}\cdots e^{\beta HS_{N}} \qquad(2)$$
Since each $S_i$ is independent of the others, we can "factorize" the $\Sigma$'s:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\left(\sum_{S_{1}\in\{-1,1\}}e^{\beta HS_{1}}\right)\cdots\left(\sum_{S_{N}\in\{-1,1\}}e^{\beta HS_{N}}\right)$$
And finally:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{j=1}^{N}\sum_{S_{j}\in\{-1,1\}}e^{\beta HS_{j}} \qquad Q.E.D.$$
Note. This result can be generalized for vectors $\vec{S}=(S_1,...,S_N)$ with components $S_i\in\{1,...,k\}$ for some integer $k$.
This is called an elementary symmetric polynomial. For your example the notation is $$e_3(x_{11}, x_{12}, x_{21}, x_{22})$$ and I imagine nobody would object if you abbreviated this to $$e_3(x_{11}\ldots x_{22})$$ or $$e_3(x_{ij}).$$
Best Answer
A variation: We denote with $[m]=\{1,2,\ldots,m\}$ and obtain \begin{align*} \color{blue}{\prod_{i=1}^n\sum_{j=1}^m a_{ij}} &=\left(a_{11}+\cdots+a_{1m}\right)\left(a_{21}+\cdots+a_{2m}\right)\\ &\qquad\cdots\left(a_{n1}+\cdots+a_{nm}\right)\\ &=\sum_{\left(j_1,\ldots,j_n\right)\in[m]^n }a_{1j_1}a_{2j_2}\cdots a_{nj_n}\tag{1}\\ &\,\,\color{blue}{=\sum_{\left(j_1,\ldots,j_n\right)\in[m]^n }\prod_{i=1}^na_{ij_i}}\\ \end{align*}
In (1) we observe that multiplication of the $n$ terms each having $m$ summands is the same as summing up all $n$-tuples $\left(j_1,\ldots,j_n\right)$ from the $n$-fold cartesion product $[m]^n$ and multiply out the elements of each $n$-tuple.