For the field:
$$ \int d^3x\phi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int\int d^3x\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=\frac{1}{\sqrt{2\omega_{p}'}}\left(a_{\vec{p}'}+a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\phi}(\vec{p}')$$
For the momentum:
$$ \int d^3x\pi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int d^3x\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=(-i)\sqrt{\frac{\omega_{p'}}{2}}\left(a_{\vec{p}'}-a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\pi}(\vec{p}')$$
So
$$\boxed{\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(\vec{p}')+i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(\vec{p}')\right)=a_{\vec{p}'}}$$
$$\boxed{\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(\vec{p}')-i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(\vec{p}')\right)=a^{\dagger}_{-\vec{p}'}}$$
Now we can use the fact that the CCR for the momentum and the field is almost the same as the CCR for the Fourier transforms:
$$[\phi(\vec{x}),\pi(\vec{y})]-i \delta^{(3)}(\vec{x}-\vec{y})=0$$
Multiplying by $e^{-i(\vec{p}\cdot\vec{x})}$ and $e^{-i(\vec{p}'\cdot\vec{y})}$ integrating with respect to $\vec{x}$ and $\vec{y}$
$$ \int d^3xd^3y\left( e^{-i(\vec{p}\cdot\vec{x})}\phi(\vec{x})e^{-i(\vec{p}'\cdot\vec{y})}\pi(\vec{y})-\pi(\vec{y})e^{-i(-\vec{p}'\cdot\vec{y})}\phi(\vec{x})e^{-i(\vec{p}\cdot\vec{x})}-i\delta^{(3)}(\vec{x}-\vec{y})\right)=0$$
We get the commutator in terms of the fields and momentum in the momentum basis:
$$[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]=(2\pi)^3i\delta^{(3)}(\vec{p}-\vec{p}') $$
so that
$$\left[a_{\vec{p}},a^{\dagger}_{-\vec{p}'}\right]=\left[\frac{1}{2}\left(\sqrt{2w_{p}}\tilde{\phi}(\vec{p})+i\sqrt{\frac{2}{w_{p}}}\tilde{\pi}(\vec{p})\right),\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(-\vec{p}')-i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(-\vec{p}')\right)\right]=\frac{1}{2}\left(-i[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]+i[\tilde{\pi}(-\vec{p}),\tilde{\phi}(-\vec{p}')] \right)=(2\pi)^3\delta({\vec{p}+\vec{p}'})$$
Final Remarks
1)I think there are some $\hbar$ missing and 2) I'm not sure about the symmetry between $\vec{p}$ and $-\vec{p}$ for the real KG field
Edit
I set $\hbar=1$, but the commutator still looks a little ugly with that $+$ sign
Edit
Given that the field is real may I say $\tilde{\phi}(\vec{p})=\tilde{\phi}^*(-\vec{p})=\tilde{\phi}(-\vec{p})$
It has to do with Feynman-Stuckelberg interpretation of negative energy solutions as outgoing antiparticles.
A plane wave mode with negative energy is given by
$$a(\vec p)e^{-i(Et-\vec p\cdot\vec x)},\quad E<0,$$
which can be written as
$$a(\vec p)e^{i(Et+\vec p\cdot\vec x)},\quad E>0.$$
Since the plane wave expansion is a sum over all 3-momenta, we can write it as
$$\int d^3pa(\vec p)e^{i(Et+\vec p\cdot\vec x)}=\int d^3pa(-\vec p)e^{i(Et-\vec p\cdot\vec x)}=\int d^3pa(\vec p)d^3pe^{ipx},\quad E>0.$$
However, all particles in the theory must have positive energy so we need to given an interpretation to negative energy/frequency modes. Changing sign of time is equivalent to changing sign of charge and reflecting 3-momentum of the particle. Hence, if we second quantize the theory and the coefficients $a,a^\dagger$ are associated to particles, then the coefficients corresponding to negative frequencies are associated to antiparticles, $b,b^\dagger$. Moreover, these modes correspond to a reflected 3-momentum (negative time) and we interpret this as outgoing antiparticles as long as we interpret the positive frequency modes as incoming particles. By outgoing we mean a particle that is created in the system and then leaves it, therefore must be associated to a creation operator. An incoming particle enters the system and is absorbed by it, thus being associated to a annihilation operator.
Best Answer
It seems there is a mistake in your $\pi (x)$ expression: there must be one minus sign near $\hat{a}^{\dagger}$.
The relation between classical and field is obvious since lagrangian (hamiltonian) of free $\varphi $ (it's not hard to see that $\varphi$ satisfies Klein-Gordon equation) field may be rewritten as lagrangian of free ossilator in terms of $\hat{\varphi} , \hat{\pi} = \dot{\hat{\varphi}}$ which have been introduced in your question. The differences between "classical" and field expressions for coordinate and momentum is caused that field in every point can be represented as set of oscillators. This follows from hamiltonian in terms of $\hat{\varphi} , \hat{\pi}$): $$ L =\frac{1}{2}\left( (\partial_{\mu}\varphi )^{2} - m^{2}\varphi^{2}\right) \Rightarrow \hat{H} = \int T^{00}d^{3}\mathbf r = \int \left(\frac{\partial L}{\partial (\partial_{0}\varphi)}\partial_{0}\varphi - L \right)d^{3}\mathbf r = $$ $$ = \frac{1}{2}\int d^{3}\mathbf r \left( m^{2}\varphi^{2} + \pi^{2} + (\nabla \varphi )^{2}\right). $$
By introducing canonical relations $$ [\hat{a}(\mathbf p), \hat{a}^{\dagger}(\mathbf p')] = \delta (\mathbf p - \mathbf p '), \quad [\hat{a}(\mathbf p), \hat{a}(\mathbf p')] = [\hat{a}^{\dagger}(\mathbf p), \hat{a}^{\dagger}(\mathbf p')] = 0 $$ you get full correspondence between "classical" and "field" operators $\hat {x}, \hat{p}$, $\hat{\varphi}, \hat{\pi}$ (up to delta-function $\delta (\mathbf x - \mathbf x ')$):
$$ [\hat{x}_{i}, \hat{p}_{j}] = i\delta_{ij}, \quad [\hat{\varphi} (\mathbf x , t), \hat{\pi}(\mathbf y, t)] = i\delta (\mathbf x - \mathbf y). $$