Mapping preserves implications Let $S$ and $T$ be sets, and let ${\phi}:S\, {\rightarrow}\,T $ be a mapping.
Let $*$ be an operation on $S$.
Let $x$ and $y$ be any elements of $S$ and let $a$ be any of the set's cancellable elements (under $*$).
Then, from the above, we have that ${\forall}x,y,a\,{\in}\,S:x\,*\,a=y\,*\,a{\implies}x=y$
From the definition of mapping this would mean that ${\phi}(x\,*\,a)={\phi}(y\,*\,a)$ and that ${\phi}(x)={\phi}(y)$.
However, my source says that not only can we deduce that:

$$[{\phi}(x\,*\,a)={\phi}(y\,*\,a)] ∧ [{\phi}(x)={\phi}(y)]$$ but also that $$[{\phi}(x\,*\,a)={\phi}(y\,*\,a)] {\implies} [{\phi}(x)={\phi}(y)]$$

How are we justified in making a jump from $∧$ to ${\implies}$?
 A: Your source doesn't mean that
$$
\forall x\,\forall y\,(\phi(x*a)=\phi(y*a) \land \phi(x)=\phi(y)),\tag{$\it\text{not generally true}$}
$$
as this implies $\forall x\,\forall y\,\phi(x)=\phi(y)$, which says that $\phi$ is a constant function.
Your source means that if $x,y$ are such that $\phi(x*a)=\phi(y*a)$, then because of the implication $\implies$, it follows that $\phi(x)=\phi(y)$, and so the conjunction follows. However, the implication involving $\phi$ is not always true.
I don't understand how you arrive at your statement:

From the definition of mapping this would mean that ${\phi}(x\,*\,a)={\phi}(y\,*\,a)$ and that ${\phi}(x)={\phi}(y)$.

This does not follow from what you've said up to that point. Counterexample to the implication and conjunction: $S=\Bbb N$, $T = \{0,1\}$, define $\phi\colon S\to T$ by:
$$
\phi(n) = \begin{cases}
0 &\text{if $n=0$,}\\
1 &\text{if $n>0$.} \end{cases}
$$
Let $*$ be $+$ on $\Bbb N$, and let $a=1$, so for all $m,n$, if $m+1=n+1$ then $m=n$. Note that $\phi(0+1) = \phi(1+1) = 1$; however, $\phi(0)\ne\phi(1)$. So, it's not true that $\forall x\,\forall y\,(\phi(x*a) = \phi(y*a) \to \phi(x)=\phi(y))$.
If $\phi$ is injective then this follows, but for an arbitrary function $\phi$ it does not. 
PS
It turns out your 'source' is  proofwiki.org/wiki/Morphism_Property_Preserves_Cancellability. The "proof" given on that page is false. There, $S,T$ both have some algebraic structure, and $\phi$ is not just any function, but a morphism — it preserves the operations. For simplicity, consider just structures with just a single binary operation. The claim for left-cancellability is:
$$
\text{If $\phi\colon S=(S,*)\to T=(T,\star)$ is a morphism, $\\$ and if $\forall x\,\forall y\,(a*x = a*y \to \phi(x)=\phi(y))$, $\\$ then $\forall x\,\forall y\,(\phi(a*x) = \phi(a*y) \to \phi(x)=\phi(y))$.}\tag{false}
$$
This is not so. Let $(S,*)=(\Bbb N, +)$ be as above. Let $T=(\Bbb N, \star)$ where $\star$ is defined as follows:
$$
a\star x = \begin{cases}
x &\text{if a=0,}\\
1 &\text{if a>0.}\\
\end{cases}
$$
Let $\phi$ be defined as above. Then $\phi(0+x) = \phi(x)=0\star\phi(x) =\phi(0)\star\phi(x)$, and if $a>0$, then $\phi(a+x) = 1 = 1\star\phi(x)=\phi(a)\star\phi(x)$. So $\phi$ is a morphism.
Just as above, however, $\phi(1+1)=\phi(2)=\phi(1)=\phi(0+1)=1$, but $\phi(0)\ne\phi(1)$.
