Unitary translation in phase space coordinate If we suppose that we can translate one point to another point in phase space $(x,p)$ through the following operators,
$$T(\Delta x) = \exp(-i p~\Delta x ) $$
and
$$T(\Delta p) = \exp(-i x~\Delta p ) ,$$
I want to see if there is any common point between these transformations and canonical transformation?
 A: Let us stick to Quantum mechanics, so, then, use P and Q for operators, and, to avoid confusion, s=Δx and t=Δp for classical shift parameters.
You then evidently wrote down the (conjugates of) celebrated U and V Weyl operators (1927), useful in the braiding form of the canonical commutation relations,
$$
e^{itQ} e^{isP} = e^{-i st} e^{isP} e^{itQ}  .
$$
Their actions on functions of x are rephase ("clock") and shift,
$$
e^{itQ}\psi(x)=   e^{itx}\psi(x),\qquad  e^{isP}\psi(x)=\psi(x+s),
$$ 
and for p,
$$
e^{isP}\phi(p)=   e^{isp}\phi(p),\qquad  e^{itX}\phi(p)=\phi(p-t).
$$ 
They have particularly handy properties and serve as a bridge to finite dimensional Hilbert spaces (clocks).
I'm not sure how much you'd like to base your classical limit from quantum canonical transformations to classical ones (it is a truly daunting and treacherous area, approached with respect, if not trepidation!) on these; but, of course, they are both canonical as they preserve Heisenberg's commutation relation, 
$$
e^{itQ} e^{isP} [Q,P] e^{-isP}e^{-itQ}= [Q,P]=i.
$$
If you really wished to get into quantum canonical transformations in phase space, where angels fear to tread, you may start from Bracken,   Cassinelli, and Wood, "Quantum symmetries and the Weyl–Wigner product of group representations", Journal of Physics A: Mathematical and General 36.4 (2003) 1033 and pursue the groaning bibliography. 
A: The connection uses the formalism of Lie transform.  Please refer to the following for additional details:


*

*Ernesto Corinadelsi, Classical Mechanics, Chap. 9

*Alex J. Dragt, Lectures on non-linear orbit dynamics
I will basically reproduce the derivation of Corinadelsi with some minor modifications. The method similar in spirit to the interaction picture of quantum mechanics. 
Suppose $H(q,p)$ is the Hamiltonian of a natural system, 
and does not depend explicitly on $t$.  Therefore, $H(q(t),p(t))=H(q_0,p_0)$ with $q_0=q(0)$ and
$p_0=p(0)$, the energy the same at any point in the motion, and
in particular is the same as the energy at the start of the motion.
The time-evolution of any function $f$ is then given by
$$
\frac{df}{dt}=\{f,H\} 
$$ 
We seek to write
$$
f(q(t),p(t))=e^{t \Omega}f(q_0,p_0)
$$
for some operator $\Omega$ with the exponential formally
defined using the series
$e^{t \Omega}:=\sum_k \frac{t^k\Omega^k}{k!}$.
For notational convenience, set
$$
z(t)=(q(t),p(t)):=z\qquad z(0)=z_0=(q_0,p_0)\, .
$$ 
Now, $z(t)$ are canonical variables, just like $z_0$, so that, 
in the Poisson bracket, we can use $\{f,H\}_{z_0}$.  
In particular, 
$$
\frac{df(z)}{dt}=\{f(z(z_0,t),H(z_0)\}_{z_0}
$$
always, and certainly
$$ 
\frac{df(z)}{dt}\Bigl\vert_{t=0}=\{f(z(z_0,t),H(z_0)\}_{z_0}\Bigl\vert_{t=0}=\{f(z_0),H(z_0)\}_{z_0}\, .
$$ 
Define
$$ 
\Omega f(z_0):= \frac{df(z)}{dt}\Bigl\vert_{t=0}=
\{f(z_0),H(z_0)\}_{z_0}\, .
$$ 
Clearly, 
$$ 
\Omega^2 f(z_0)=\Omega(\Omega f(z_0))=\frac{d^2 f}{dt^2}
\Bigl\vert_{t=0}=\{\{f(z_0),H_0\},H_0\}_{z_0}
$$ 
etc.  In particular, assuming $f(z(t))$ is smooth, it has a series
expansion
\begin{align}  
f(z(t))&=\sum_{n=0}^{\infty}\frac{t^n}{n!}\left(\frac{d^n f(z(t))}{dt^n}
\right)_{t=0}=\sum_{n=0}^{\infty}\frac{t^n}{n!}\Omega^n f(z_0)=e^{t\Omega}f(z_0)
\end{align} 
Now, $\Omega$ is determined only by $H$. For instance, if your Hamiltonian is
$H=p$, the generator of translations, we have, for any function:
$$
\Omega f(z_0)=\{f(z_0),H(z_0)\}=\frac{\partial f}{\partial q_0}
\frac{\partial p_0}{\partial p_0}-\frac{\partial f}{\partial p_0}
\frac{\partial p_0}{\partial q_0}=\frac{\partial f}{\partial q_0}
$$ 
and, since the function $f$ is arbitrary, we find
$$
\Omega=\frac{\partial }{\partial q_0}\, ,\qquad
e^{t\alpha\Omega}f(q_0,p_0)=f(q_0+\alpha t,p_0)\, ,
$$
which can easily be verified using any $q_0^k$.
If, on the other hand, $H=q$, then one find
$$
\Omega=-\frac{\partial }{\partial p_0}\, .
$$
