Is it possible that $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ same as $a^{2} - 1 \equiv 0 \pmod{m}$? $a^{2} - 1 \equiv 0\pmod{m\cdot a}$ same as $a^{2} - 1 \equiv 0 \pmod{m}$?
Given that $a$ and $m$ are relatively prime, is it possible for the above equation to hold true? My rational behind this is the following:
Because $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$,  this implies that $m\cdot a\mid \left(a^{2} - 1\right)$, which implies $a^{2} - 1 = l\cdot m\cdot a, l \in \mathbb Z$ 
Is it wrong to then say that: let $t = l\cdot a, t \in \mathbb Z$. Hence, we have $a^{2} - 1 = t\cdot m$, thus implying that $a^{2} - 1 \equiv 0 \pmod {m}$
Note that $a$ is a positive integer in this case
EDIT: What I meant to ask was does $a^{2} - 1 \equiv 0 \pmod{m\cdot a}$ imply $a^{2} - 1 \equiv 0 \pmod{m}$?
 A: It is not the same:  it only goes in one direction. Yes, it is possible they are both true, but not a universal certainty.  And, I don't think it has much to do with $a^2-1$.  
If $x \equiv 0 \bmod ma$ then $ma \mid x$. Since clearly $m \mid ma$ we have $m \mid x$ and so $x \equiv 0 \bmod m$.  For your case, let $x=a^2-1$, but this really doesn't matter:  it's still true. 
The other direction may be false.  For instance $10 \equiv 0 \bmod 5$ but $10$ is not congruent to $0 \bmod 15$.  Or, take my example from my comment with $a=4$ and $m=3$:  $15\equiv 0 \bmod 3$ yet $15$ is not congruent to $0 \bmod 12$.  
Edit: I've interpreted your question "are they the same" as asking "is there an if and only if in between."
A: This statement is not biconditional. 
To be precise, 
if $a^2-1\equiv 0\mod (m*a)$, then $a^2-1\equiv 0\mod a$ .
if $a^2-1\equiv 0\mod (a)$, then it is not always true that $a^2-1\equiv 0\mod (m*a)$.
Your logic is correct, however some constraints may be removed. More generally, $m$ and $a$ do not have to be coprime,it does not have to be a, and the exponent can be any integer $n$, not just $2$. 
A more general statement is as follows: 
If $a^n-1\equiv 0\mod N$, then $a^n-1\equiv 0\mod f$, for every factor $f$ of $N$.
Proof. We can see that $N|a^n-1$. Therefore all of its factors must too. $f|a^n-1\Longleftrightarrow a^n-1\equiv 0\mod f$.
I suggest looking into the concept of multiplicative order for a more thorough understanding. 
A: Of course $\,ma\mid a^2-1 \,\Rightarrow\,  m\mid a^2-1\,$ is always true, $ $ by $\ (ma)n = m(an)$.
The converse is true $\iff m = 0.\,$ For if $\,m=0\,$ both become sides are identical: $\,0\mid a^2-1$.
And if $\,m\neq 0\,$ then the RHS: $\,m\mid a^2-1\,$ holds for $\,a = 1+km \equiv 1\pmod{\!m}.\,$
But the LHS $\,\Rightarrow\, a\mid a^2-1\,\Rightarrow\, a\mid 1,\,$ falsified by choosing $\,k\,$ so $\,a = 1+km > 1.$
A: $a^2 -1 \pmod {ma} \implies ma|a^2 - 1 \implies m|a^2-1$ so that one direction always holds if it ever holds.
However $ma|a^2 - 1 \implies \exists k: a^2 - 1 = mak$ so $ a -\frac 1a =mk\in \mathbb Z$ which implies $a \equiv \pm 1$ so $a^2 - 1 \pmod {ma}$ only if $a = \pm 1$.
So, yes,  IF $a^2 -1 \pmod {ma}$ then $a^2 -1 \pmod m$.  BUT $a^2 - 1 \pmod{ma}$ only when $a = \pm 1$.
The other way:
$a^2 -1 \equiv \pmod m \implies a^2 - 1\equiv km \pmod {am}$ for some integer $k: 0\le k< m$ (and chinese remainder theorem says that such a $k$ is unique) but there is no reason to assume $k =0$ (and lots of reason to assume it isn't).
So so any $a^2 -1 \pmod {m}$ where $a\not \equiv \pm 1 \pmod m$ will be a counter example.
Fo instance  $5^2 -1 \equiv 0 \pmod {24}$  and $\gcd(a,5) = 1$ but $5^2 -1 \not \equiv 0 \pmod {5*24 = 120}$ 
