The ideal generated by homogeneous elements of a graded algebra is graded

Let $J$ consist of all finite sums ${\sum }_{k=1}^m{a}_k{s}_k{b}_k$, with $a_k,b_k\in A$ and $s_k\in S$, including the empty sum $0$. Concatenating sums and replacing $a_k$ by $-a_k$ show that $J$ is an additive subgroup; multiplying a sum on the left or right absorbs that factor into $a_k$ or $b_k$, using associativity from [F1], so [F2] makes $J$ a two-sided ideal. It contains every $s\in S$ as $1s1$. Every two-sided ideal containing $S$ contains each $a_k{s}_k{b}_k$ and every such finite sum. Thus $J$ is the smallest two-sided ideal containing $S$, and [F3] gives $I=J$.

For each summand $asb$ in a finite expression defining $J$, write $a=a_{\bar{0}}+a_{\bar{1}}$ and $b=b_{\bar{0}}+b_{\bar{1}}$. Then $asb={\sum }_{i,j\in \{0,1\}}{a}_{\bar{i}}s{b}_{\bar{j}}$; each term is homogeneous by [F1] and lies in $J$. Therefore every homogeneous component of every element of $J$ lies in $J$, proving $I=(I\cap A_{\bar{0}})\oplus (I\cap A_{\bar{1}})$. Every coset in $A\mathrm{/}I$ decomposes into its even and odd component classes; if a class lies in their intersection, its two homogeneous representatives differ by an element of the graded ideal $I$, so both representatives lie in $I$ and the class is zero. The quotient grading is direct, and the product respects it because the product in $A$ does.