Definition of Tensor Product for Scalar Extension In Dummit and Foote p360, they constructed special case of tensor product for scalar extension. $R$ is a subring of $S$; we have $R$-module $N$, and we aim to construct a $S$-module $S \otimes _R N$:

(i) Construct the  free abelian group on set $S \times N$. i.e. formal linear combinations of $(s,n)$. 
(ii) Quotient group, $S \otimes_{R} N$, defined by qoutienting out subgroup generated by elements $$ (s_1+ s_2, n) - (s_1, n) - (s_2,n), \quad (s,n_1+n_2) - (s,n_1)- (s,n_2), \quad (sr,n) - (s,rn).$$
where $s,s_1,s_2 \in S$, $n,n_1,n_2 \in N$ and $r \in R$, where $rn$ refers to $R$-moudle structure already in $N$.
(iii) Define an element of quotient group by $s \otimes n$. Then the $S$-module $S \otimes_{R} N$ is given by 
$$ s( \sum s_i \otimes n_i ) = \sum (ss_i) \otimes n_i .$$
and there is a $R$-homomorphism $i:N \rightarrow S \otimes_R N$ given by $n \mapsto 1 \otimes n$. 

Question Regarding Choice of Quotient Group: A: Assuming we don't need the existence of $i:N \rightarrow S \otimes_R N $ in (iii): Is quotienting out the subgroup generated by elements of form $$(s_1+s_2,n) - (s_2,n)-(s_2,n)$$ enough to make $S \times N$ an $S$-module from action $\sum s_i \otimes n_i \mapsto \sum ss_i \otimes n_i $? 
B: In order for $i$ to be a $R$ homomorphism, we only need $1 \otimes (n_1+n_2) = 1 \otimes n_1 + 1 \otimes n_2 $ and $1 \otimes rn  = r (1 \otimes n) $ for all $r \in R, n \in N$. 
So isn't quotienting out subgroup generated by
$$(s_1+s_2,n) - (s_1, n) -(s_2,n), \quad (1, n_1+n_2) - (1,n_1) - (1,n_2), \quad (r,n) - (1,rn) .$$
enough to satisfy all conditions? 

EDIT (Some thoughts): In retrospect, I noticed in order for the map $\sum s_i \otimes n_i \mapsto \sum (ss_i) \otimes n_i $ to be well-defined we need, with $H$ the subgroup which we quotient out from, 
$$ \sum (s_i, n_i) \in H  \Rightarrow \sum (ss_i, n_i)  \in H $$
So elements of subgroup $H$ must be closed under action of elements from $S$. Conversely, if this condition holds, our map is well-defined, for $ \sum s_i \otimes n_i = \sum s_i' \otimes n_i'$, yields
$$ \sum (s_i, n_i) - \sum (s_i' , n_i') \in H \Rightarrow \sum (ss_i , n_i) - \sum (ss_i' , n_i') \in H \Rightarrow \sum ss_i \otimes n_i = \sum ss_i' \otimes n_i' $$
Is this interpretation correct?  
