Quantization of complex scalar field

I am afraid what is wrong is in your initial writing.

A complex scalar field should be written as
$phi(x) = int frac{d^3p}{(2 pi)^3} frac{1}{sqrt{2 omega_p}} (a_p e^{-ipx} + b_p^dagger e^{ipx})$
And, by complex conjugation,
$phi^*(x) = int frac{d^3p}{(2 pi)^3} frac{1}{sqrt{2 omega_p}} (a_p^dagger e^{ipx} + b_p e^{-ipx})$
Clearly $a_p^dagger neq b_p^dagger$ as these operators create particles of opposite charge. However in both cases $omega_p = sqrt{vec p^2 +m^2} gt 0$.

To get the normal ordering of the $b's$ operators, at the end of the demonstration you have to apply the equal-time commutation relations
$[b_p, b_{p'}^dagger] = (2 pi)^3 delta^3 (vec p - vec p')$

Note: Your notation of the $b$ operator as creator is confusing. The notation used in literature for a complex scalar field and reported here is a generalization of the notation applied to a real scalar field.

To begin with, Peskin gives the following action: $$ mathcal{S} = int d^4 left[ partial_{mu} phi^{ast} partial^{mu} phi - m^2 phi^{ast} phi right]. $$ Let's begin by considering the classical field theory. The equation of motion is $$ left( partial^2 + m^2 right) phi = 0 $$ (and its complex conjugate). A general solution to this equation is given by $$ phi(mathbf{x},t) = int frac{d^3 p}{(2 pi)^3} frac{1}{sqrt{2 E_mathbf{p}}} left( a_{mathbf{p}} e^{i mathbf{p} cdot mathbf{x} + i E_{mathbf{p}} t} + b_{mathbf{p}} e^{i mathbf{p} cdot mathbf{x} - i E_{mathbf{p}} t} right), $$ where $E_{mathbf{p}} = sqrt{|mathbf{p}|^2 + m^2}$. At this point, the quantities $a_{mathbf{p}}$ and $b_{mathbf{p}}$ are just some complex functions of momentum. We can also write down the classical Hamiltonian of this system, where the conjugate momentum is $$ pi(mathbf{x},t) = partial_t phi^{ast}(x,t) = - i int frac{d^3 p}{(2 pi)^3} sqrt{frac{E_mathbf{p}}{2}} left( a_{mathbf{p}}^{ast} e^{-i mathbf{p} cdot mathbf{x} - i E_{mathbf{p}} t} - b_{mathbf{p}}^{ast} e^{-i mathbf{p} cdot mathbf{x} + i E_{mathbf{p}} t} right) $$ (and its complex conjugate).

Now we go to the quantum theory, where the fields $phi$ and $pi$ (and therefore the coefficients $a_{mathbf{p}}$ and $b_{mathbf{p}}$) become operators. The starting point of quantization is the equal-time canonical commutation relations: $$ [phi(mathbf{x},t),pi(mathbf{y},t)] = [phi^{ast}(mathbf{x},t),pi^{ast}(mathbf{y},t)] = i delta^3(mathbf{x} - mathbf{y}) $$ We cannot assume at this point that $a_{mathbf{p}}$ and $b_{mathbf{p}}$ are bosonic annihilation operators, we need to actually calculate their commutation relations from the above equations. Inverting the above formulae, we have $$ a_{mathbf{p}} = e^{- i E_{mathbf{p}} t} int d^3 x , e^{- i mathbf{p} cdot mathbf{x}} left( sqrt{frac{E_{mathbf{p}}}{2}} , phi(mathbf{x},t) - i frac{1}{sqrt{2E_{mathbf{p}}}} , pi^{ast}(mathbf{x},t) right), $$ $$ b_{mathbf{p}} = e^{- i E_{mathbf{p}} t} int d^3 x , e^{- i mathbf{p} cdot mathbf{x}} left( sqrt{frac{E_{mathbf{p}}}{2}} , phi(mathbf{x},t) + i frac{1}{sqrt{2E_{mathbf{p}}}} , pi^{ast}(mathbf{x},t) right). $$ Now we can compute the commutators directly. They are $$ [a_{mathbf{p}},a^{dagger}_{mathbf{p}'}] = - (2 pi)^3 delta^3(mathbf{p} + mathbf{p}'), qquad [b_{mathbf{p}},b^{dagger}_{mathbf{p}'}] = (2 pi)^3 delta^3(mathbf{p} + mathbf{p}'). $$ As you can see, one of these commutators is the wrong sign compared to the usual bosonic commutation relations, and as a result the construction of the usual bosonic Fock space proceeds differently. (In particular, $[a^{dagger} a, a^{dagger}] propto - a^{dagger} $ so $a^{dagger}$ is actually a lowering operator.) To get the usual commutation relations, we should define $$ tilde{a}_{mathbf{p}} = a^{dagger}_{mathbf{p}}, quad tilde{a}^{dagger}_{mathbf{p}} = a_{-mathbf{p}}, quad tilde{b}_{mathbf{p}} = b_{mathbf{p}}, quad tilde{b}^{dagger}_{mathbf{p}} = b^{dagger}_{-mathbf{p}}. $$ This gives us the usual result. In textbooks they often write down the initial expansion in such a way so that the answer comes out correctly, but in practice one needs to go through the above procedure (until you've done it enough times that it has become automatic).