Problem
Let $R'$ and $R''$ be (commutative unitary) rings. Consider $R := R' \times R''$ and set $S := \{(1, 1), (1, 0)\}$. Prove $R' = S^{-1}R$. (By the convention of the book, I believe the exercise wants to find an isomorphism $R' \simeq S^{-1}R$.)
(This problem is from "A Term of Commutative Algebra" section 11.5 [number might differ] by Allen Altman and Steven Kleiman.)
Given Solution
Let's show that the projection map $\pi: R' \times R'' \rightarrow R'$ has the UMP. First note that $\pi S = \{1\} \subset R'^\times$. Let $\psi: R' \times R'' \rightarrow B$ be a ring map such that $\psi(1, 0) \in B^\times$. Then in B,
$$ \psi(1, 0) \cdot \psi(0, x) = \psi((1, 0) \cdot (0, x)) = \psi(0, 0) = 0 \text{ in } B \text{.} $$
Hence $\psi(0, x) = 0$ for all $x \in R''$. So $\psi$ factors uniquely through $\pi$.
My Issue
I don't understand the solution. When I was trying on my own, I knew that I should use the UMP. Indeed, since for the projection map $\pi: R' \times R'' \rightarrow R'$ it is $\pi(S) = \{1\}$ the condition is fulfilled. There is a unique map $\rho: S^{-1}R \rightarrow R'$ with $\rho \circ \varphi_S = \pi$ where $\varphi_S: R \rightarrow S^{-1}R$ and $\varphi_S(x) := x/1$. We want to show that $\rho$ is an isomorphism. But (and here must be my mistake) $\ker(\rho) = \{(0, y) | y \in R''\}$ and $\rho$ cannot be injective.
I would appreciate it if somebody could clear up my misunderstanding. Thank you.
The issue with your issue is that $j(\{0\} \times R'') = (0,0)/(1,1)$, so your expression for $\ker(\rho)$ is wrong.
Let $W := \{1\}$. We have $W^{-1}R' \cong R'$ (we'll just identify them). Consider $\iota : R' \rightarrow S^{-1}R$ given by $\iota(x) := (x,0)/(1,1)$. Notice that $\iota(1) = (1,0)/(1,1)$ which is a unit by definition. It turns out studying this map along with the maps you defined does the trick by the following commutative diagram:
That's not super illuminating, so let's break it down. First, we get $\rho_S$ from your argument. Next, by identfiying $W^{-1}R'$ and using $\iota$, we can get a unique map $\rho_W : R' \rightarrow S^{-1}R$. Now we claim that $\rho_W \circ \rho_S$ is the identity. By composing maps, we get a homomorphism $\iota \circ \pi : R \rightarrow S^{-1}R$ and so by the universal property this must give us a unique homomorphism $\rho' : S^{-1}R \rightarrow S^{-1}R$ such that $\rho' \circ j = \iota \circ \pi$. We claim that $j = \iota \circ \pi$, forcing $\rho' = \text{Id}$. This follows if we can show $$(x,y)/(1,1) = (x,0)/(1,1).$$ Notice that $$ (1,0)((1,1)(x,y) - (1,1)(x,0)) = (0,0),$$ so the result follows.
Next, we claim that $\rho_W \circ \rho_S \circ j = \iota \circ \pi$. Notice that $$ \rho_W \circ \rho_S \circ j = \rho_W \circ \pi = \rho_W \circ \text{Id} \circ \pi = \iota \circ \pi.$$
By uniqueness this forces $\rho_W \circ \rho_S = \text{Id}$, and so $\rho_S$ has a left inverse. This forces $\rho_S$ to be injective, and clearly $\rho_S$ is surjective, so we win.
Alternatively, we can do the same kind of argument to show that $\rho_S \circ \rho_W = \text{Id}$, and we can get that $\rho_S$ is an isomorphism using just the universal property.