Okay...I think this answer finally takes care of the difficulty. Thanks to arctictern for all his help :)
Let $g \in (G_{1}\times G_{2})^{\prime}$. Then, since $(G_{1} \times G_{2})^{\prime}$ is the subgroup generated by all commutators of elements of $G_{1} \times G_{2}$, $g$ is the product of commutators of elements of $G_{1} \times G_{2}$. Letting $(a_{i},b_{i})$, $(c_{i},d_{i})\in G_{1} \times G_{2}$, we have that
$\begin{align} g = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})]\,\text{for some}\, t, \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = ([a_{1},c_{1}]\cdot [a_{2},c_{2}]\cdot \,\, \cdots \, \, \cdot [a_{t},c_{t}], \,[b_{1},d_{1}]\cdot [b_{2},d_{2}]\cdot \,\, \cdots \, \, \cdot [b_{t},d_{t}]) \in G_{1}^{\prime} \times G_{2}^{\prime}\end{align}$
So, we have the inclusion $\mathbf{(G_{1}\times G_{2})^{\prime}\subseteq G_{1}^{\prime}\times G_{2}^{\prime}}$
In the other direction, let $h \in G_{1}^{\prime} \times G_{2}^{\prime}$.
To belong in $ G_{1}^{\prime} \times G_{2}^{\prime}$, $h$ must be the direct product of an element of $G_{1}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{1}$, and an element of $G_{2}^{\prime}$, which is the (group operation) product of commutators of elements of $G_{2}$.
Let $h_{1}$ be such an element of $G_{1}^{\prime}$ and let $h_{2}$ be such an element of $G_{2}^{\prime}$, and let $h = h_{1} \times h_{2}$. Then,
$h_{1} = [a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}]$ for some $t$, where each $a_{i}$, $c_{i}$ is an element of $G_{1}$
and $h_{2} =[b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}]$ for some $t$, where each $b_{i}$, $d_{i}$ is an element of $G_{2}$.
So,
$\begin{align}h = h_{1} \times h_{2} = ([a_{1},c_{1}]\cdot [a_{2}, c_{2}] \cdot \, \, \cdots \, \, \cdot [a_{t},c_{t}], [b_{1},d_{1}]\cdot [b_{2}, d_{2}] \cdot \, \, \cdots \, \, \cdot [b_{t},d_{t}])\\ =(a_{1}^{-1}c_{1}^{-1}a_{1}c_{1}a_{2}^{-1}c_{2}^{-1}a_{2}c_{2}\cdots a_{t}^{-1}c_{t}^{-1}a_{t}c_{t}, \,b_{1}^{-1}d_{1}^{-1}b_{1}d_{1}b_{2}^{-1}d_{2}^{-1}b_{2}d_{2}\cdots b_{t}^{-1}d_{t}^{-1}b_{t}d_{t}) \\ = (a_{1}^{-1},b_{1}^{-1}) (c_{1}^{-1},d_{1}^{-1})(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2}^{-1},b_{2}^{-1})(c_{2}^{-1},d_{2}^{-1})(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t}^{-1},b_{t}^{-1})(c_{t}^{-1},d_{t}^{-1})(a_{t},b_{t})(c_{t},d_{t}) \\ = (a_{1},b_{1})^{-1}(c_{1},d_{1})^{-1}(a_{1},b_{1})(c_{1},d_{1})\cdot (a_{2},b_{2})^{-1}(c_{2},d_{2})^{-1}(a_{2},b_{2})(c_{2},d_{2})\cdot \, \, \cdots \, \,\cdot(a_{t},b_{t})^{-1}(c_{t},d_{t})^{-1}(a_{t},b_{t})(c_{t},d_{t}) \\ = [(a_{1},b_{1}),(c_{1},d_{1})]\cdot [(a_{2},b_{2}),(c_{2},d_{2})]\cdot\,\, \cdots \,\, \cdot[(a_{t},b_{t}),(c_{t},d_{t})] \in (G_{1}\times G_{2})^{\prime} \end{align}$
So, we have the inclusion $\mathbf{G_{1}^{\prime}\times G_{2}^{\prime} \subseteq (G_{1}\times G_{2})^{\prime}}$.
Thus, since we have inclusion in both directions, we have established the equality $\mathbf{(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime}\times G_{2}^{\prime}}$ for the right things this time, hopefully.
Best Answer
The statement says that $G'$ equals the normal closure N of the set of commutators of the elements of $S$, which is the smallest normal subgroup containing the set of such commutators. To prove this, divide G by N. The quotient group is abelian since it's generators commute. Thus, $N$ contains $G'$. It is also clearly contained in $G'$. Hence, they are equal.