Is the language A decidable?
I don't understand how you call $R$ from $S$. The following is a possible solution. 

Let's reduce $A_{TM}$ (halting problem) to  $\overline{A} = \{ \langle  M \rangle | M \text{ is } TM \text{ such that } L(M) = \Sigma^*  \}$ instead of $A$, since decidability of $\overline{A}$ would imply decidability of $A$.

First note that $M_\emptyset (\text{ TM accepting nothing )} \notin$ $\overline{A}$ and it is clearly that $\overline{A}$ is not empty since a TM machine simply accepting any string is in $\overline{A}$, so let's choose $M_1$ from the set $\overline{A}$ and let $L_1$ be the language recognizable by $M_1$.

For any $\langle M, w \rangle$ construct a new TM $M'$

  M'(x)
   Simulate M on w
   If M accepts w then run M_1 on x
   If M_1 accepts x then ACCEPT


Now let's show the reduction:
$\langle M,w \rangle \in A_{TM} \Rightarrow M \text{ accepts } w \Rightarrow M'$ accepts $x$ exactly when $M_1$ accepts $x \Rightarrow L(M') = L_1$ which implies $\langle M_1 \rangle \in \overline{A}$. So we have obtained 

$$\langle M,w \rangle \in A_{TM} \Rightarrow \langle M' \rangle \in \overline{A}$$

Similarly, $\langle M,w \rangle \notin A_{TM} \Rightarrow M \text{ does not accepts } w \Rightarrow M' \text{ accepts no string } \Rightarrow L(M') = \emptyset $ which implies $\langle M' \rangle \notin \overline{A}$. So we have obtained
$$\langle M,w \rangle \notin A_{TM} \Rightarrow \langle  M' \rangle \notin \overline{A}$$

Now if $\overline{A}$ were decidable then on any input $\langle M, w \rangle$ we could construct $M'$ as above and decide if $M$ halts on $w$ by checking $\langle M' \rangle \in \overline{A}$.

Finally, the general case of the proof above is known as the Rice's theorem, which you could apply directly to prove that $A$ is not decidable.