$ℏ$ in the canonical commutation relation I am wondering what the physical meaning of the introduction of a "new" constant $\hbar$ in the CCR $[\hat{x},\hat{p}]=i\hbar$ is if you compare it to the classical Poisson-bracket $\{x,p\}=1$?
I understand that it has to be imaginary since both $\hat{x}$ and $\hat{p}$ are hermitian operators, but isn't it unitless as in classical mechanics since $x$ and $p$ are still describing the position and momentum of a system?
Also, how is it the quantization of energy introduced with $E=\hbar\omega$ into this equation?
 A: Note that while the commutator $[\cdot,\cdot]$ is dimensionless, the Poisson bracket $\{\cdot,\cdot\}$ carries dimension of inverse angular momentum. So a quantity of dimension of angular momentum is needed in the CCR for dimensional reasons alone.
A: Note: although the dimensional analysis in this answer is easier to keep track of if all canonical coordinates in the Lagrangian are of the same dimension, with care it can be generalized.

but isn't it unitless as in classical mechanics

Recall that the Poisson bracket $\{f,\,g\}=\nabla_qf\cdot\nabla_pg-\nabla_pf\cdot\nabla_qg$ has the dimension of $fg/(q\cdot p)$ and hence $fg/S$ (since $p=L_{\dot q}$ has the same dimension as $S_q$) rather than $fg$, whereas $[f,\,g]=fg-gf$ has the dimension of $fg$, and therefore we must introduce a factor $\hbar$ with the dimension of $S$.

how is it the quantization of energy

Plane-wave solutions $\exp i(k\cdot x-\omega t)$ obtain a generator $i\partial_t$ of infinitesimal time translations, $i\partial_t$, plus a generator $-i\nabla$ of infinitesimal space translations. But if these generators are defined with an extra $\hbar$ factor, their dimensions are those of energy and momentum, which respectively complement time viz. the Beltrami identity and canonical coordinates viz. the definition of conjugate momenta.
As for quantization of energy, if you mean that in the sense of discretization rather than operator eigenequations, the fact that discrete energy-$\hbar\omega$ photons are a suitable model of certain empirical phenomena comes down to its implying a certain law, or a later, more general context.
A: Dirac, in his Principles of Quantum Mechanics, showed that "at any rate for the simpler ones", the quantum bracket, i.e. the commutator, should be a multiple of the classical Poisson bracket: 
In Eq.(7), $[u,v]$ is actually the Poisson bracket, whereas the LHS is the commutator (unusual notation by modern practice).
The text makes it clear that Dirac chooses the multiple to be $i\hbar$ based on experiment and dimensionality considerations.  Indeed simple choice
$$
\{A,B\}_{PB}\to  \frac{1}{i\hbar}
\left(\hat A\hat B-\hat B\hat A\right) \tag{1}
$$
(or rather searching for operators $\hat A$ and $\hat B$ so Eq.(1) always holds)
is known to eventually lead to inconsistencies: this is Groenewold's theorem.  There is an excellent discussion of this in

Chernoff, Paul R. "Mathematical obstructions to quantization." Hadronic J.;(United States) 4.CONF-8008162- (1981).

but the paper is actually quite hard to find.
A: To add to the other answers, you can do classical mechanics with a Poisson bracket that is some positive real number that is $\neq 1$ (but also $\neq 0$). Let's say you use $ \{x, p\}=a$.
$$\frac{df}{dt}=\{f, H\}$$
To evaluate this, you would use the product rule for Poisson brackets $\{fg, h\}=f\{g, h\}+\{f, h\}g$ combined with $\{x, p\}=a$. The final result will be the same as if you used the usual un-modified Poisson bracket, but multiplied your answer by $a$.
This formulation expresses the same physics. You can scale the unit of time on the LHS (=$\frac{df}{dt}$) to compensate for the extra constant. Or you can modify the evolution law to:
$$\frac{df}{dt}=\frac{1}{a} \{f, H\}$$
which is reminiscient of the presence of $\hbar$ in the Heisenberg equation of operator evolution.
To answer your other question, $\hbar$ represents the scale at which Quantum effects become significant. When the action is far greater than $\hbar$, the path integral favors the classical path. So, Quantum effects are suppressed at scales far greater than $\hbar$
$\hbar$ also shows up in the "step-size" for discretized observables. This anticipates the fact that when the relevant observables are valued far greater than $\hbar$, their spectrum almost becomes continuous, like in classical physics.
But why would the scale at which Quantum effects become significant be the same as the constant that appears in $[X, P]=i\hbar$?
Naively, this is because $[X, P]\rightarrow 0$ as $\hbar \rightarrow 0$. So, $X$ and $P$ become almost commuting at scales in which $\hbar$ is negligible.
More rigorously, you can derive things like the uncertainty principle and the path integral by taking $[X, P]=i\hbar$ as your starting point. These results show that $\hbar$ represents the scale at which Quantum mechanics becomes relevant.
