Showing a map from $\Bbb Q$ to the field of fractions of $\Bbb Z 1_{R}$ is well defined. Suppose that I have a field $R$ with characteristic $0$. Let $Q= \{uv^{-1} | u,v \in \Bbb Z 1_{R}, v \neq 0 \}$. I want to show that the map $f :\Bbb Q \rightarrow Q$ defined as $f(n/m) = (n1_{R})(m1_{R})^{-1}$ is well defined.

Suppose that $\frac {a}{b} = \frac{n}{m}$, then $am-nb = 0$ and 
$f(n/m)=(n1_{R})(m1_{R})^{-1}$ which I need to show is equal to 
$(a1_{R})(b1_{R})^{-1}$

Here I am not sure how to proceed. Hints appreciated.
To show that $f$ is well-defined, let $\frac{a}{b}=\frac{m}{n}$ where $b,n \neq 0$. Then $an-mb=0$. Since $\mathrm{char}R=0$, the homomorphism $\varphi:\mathbb{Z} \to R$ sending $1 \mapsto 1_R$ is injective. 

In fact, you can show  that for any field $F$, $\mathrm{char}F=m$ if and only if $\ker\{\varphi: \mathbb{Z} \to F\}=m\mathbb{Z}$. Sometimes this is taken as the definition of $\mathrm{char}F$.

Now consider $\varphi(an-mb)=0$. Since $f$ is a ring homomorphism, we can write

$$\varphi(a)\varphi(n)=\varphi(m)\varphi(b)$$
Injectivity of $\varphi$ implies that $\varphi(b),\varphi(n) \neq 0$. Hence,
$$\frac{\varphi(a)}{\varphi(b)}=\frac{\varphi(m)}{\varphi(n)}$$

Using your notation, $\varphi(t)=t1_R$. Hence, the equality above is the same as

$$f(\frac{a}{b})=\frac{a1_R}{b1_R}=\frac{m1_R}{n1_R}=f(\frac{m}{n})$$
Hence, $f: \mathbb{Q} \to Q$ is well-defined.

More generally, suppose that $f: D_1 \to D_2$ is an injective ring homomorphism between two integral domains $D_1$ and $D_2$. Denote their fields of fractions by $K_1$ and $K_2$, respectively. Then $f$ extends to a ring homomorphism $\hat{f}: K_1 \to K_2$ given by $\frac{a}{b} \mapsto \frac{f(a)}{f(b)}$. The proof is exactly similar and I leave it to you. In your question, $D_1=\mathbb{Z}$ and $D_2=R$. Injectivity of $f$ comes from $\mathrm{char}R=0$. That's all.
$l:=(n1_R)(m1_R)^{-1}$

so 

$(m1_R) l=n1_R$

then 

$(b1_R) l= (n1_R)^{-1}(nb1_R) l= $

$(n1_R)^{-1}(am1_R)l=a1_R(n1_R)^{-1}(n1_R)l=a1_R$

To sum up

$l=a1_R(b1_R)^{-1}$