You can find the transformation law for $u^s(p)$ by demanding that the spinor field transform as
$\psi(x)\rightarrow \psi'(x')=U^{-1}(\Lambda)\psi(x')U(\Lambda)=\Lambda_{1/2}\psi(x)$.
You already know how the creation / annihilation operators transform from the condition that the 1-particle states transform correctly and you can then find the correct transformation law for $u^s(p)$. Then, armed with this transformation law you can do the transformation in the opposite direction (which is what Peskin and Schroeder do) and you get their result.
In particular, we have
$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}U^{-1}(\Lambda)a^{s\dagger}_pU(\Lambda)u^s(p)e^{ip.\Lambda x} + $similar terms
where I've ignored the summation and the other operator since its analogous to this.
Changing the dummmy variable $p$ to $\Lambda p$ we get
$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3\Lambda p}{(2\pi)^3}\frac{1}{\sqrt{2E_\Lambda p}}U^{-1}(\Lambda)a^{s\dagger}_{\Lambda p}U(\Lambda)u^s(\Lambda p)e^{ip. x}$
since $(\Lambda p)(\Lambda x)$=$px$
We also have $U^{-1}(\Lambda)a^{s\dagger}_{\Lambda p}U(\Lambda)=\sqrt{\frac{2E_p}{2E_{\Lambda p}}}a^{s\dagger}_{p}$ giving us
$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3\Lambda p}{(2\pi)^3}\frac{\sqrt{2E_p}}{2E_\Lambda p}a^{s\dagger}_{ p}u^s(\Lambda p)e^{ip. x}$
The measure is Lorentz invariant so we can rewrite it as
$U^{-1}(\Lambda)\psi(x')U(\Lambda)=\displaystyle\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}a^{s\dagger}_{ p}u^s(\Lambda p)e^{ip. x}$
Now we demand that this equals
$\Lambda_{1/2}\psi(x)=\Lambda_{1/2}\displaystyle\int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}a^{s\dagger}_{ p}u^s(p)e^{ip. x}$
and we immediately see that we must have
$u^s(\Lambda p)=\Lambda_{1/2}u^s(p)$.
Now you can apply the inverse transformation,
$\psi(x)\rightarrow U(\Lambda)\psi(x)U^{-1}(\Lambda)$ to get the result Peskin & Schroeder have.
The first statement
$| p \rangle$ is an eigenstate of $\hat{P}^{\mu}$ operators with eigenvalue $p^{\mu}$. It's natural then that
$$
U(\Lambda )| p \rangle
$$
is also eigenstate of $\hat{P}^{\mu}$:
$$
\hat{P}^{\mu}U(\Lambda )| p \rangle = U(\Lambda )\left[ U(\Lambda^{-1})\hat{P}^{\mu}U(\Lambda )\right]| p \rangle = U(\Lambda )\Lambda^{\mu}_{\ \nu}\hat{P}^{\nu}| p \rangle = \Lambda^{\mu}_{\ \nu}p^{\nu}U(\Lambda )| p \rangle .
$$
In general (for non-scalar case when $| p \rangle = | p ,\sigma \rangle$) from this follows that
$$
U(\Lambda )| p, \sigma \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \Lambda p, \sigma {'}\rangle
$$
and for all states $| p, \sigma \rangle $ states $U(\Lambda )| p, \sigma \rangle$ also belongs to the Hilbert space of eigenstates, so if in the beginning $| p, \sigma \rangle $ belongs to some definite orbit of the Lorentz group, then $U(\Lambda )| p, \sigma \rangle$ will also belong to this orbit:
$$
| p, \sigma \rangle = |\mathbf p , \sigma \rangle_{p_{0} = f(\mathbf p , m)} \Rightarrow U(\Lambda )| p, \sigma \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \mathbf {\Lambda p}, \sigma {'}\rangle_{p_{0} = f(\mathbf p , m)}.
$$
The invariance condition of $\langle p | q \rangle $ (since $\langle p| q\rangle = C(p) \delta (\mathbf p - \mathbf q) $ lorentz-invariance condition means that $C(p) \sim E_{p})$ means that $C_{\sigma \sigma {'}}$ satisfies $C_{\sigma \sigma{''}}C^{\dagger}_{\sigma {''}\sigma{'}} = \delta_{\sigma \sigma {'}}$. In simplest scalar case this directly leads to
$$
\tag 1 U(\Lambda )| \mathbf p \rangle = e^{i\alpha}| \mathbf {U(\Lambda ) p}\rangle ,
$$
where $\alpha$ is some parameter of transformation given by $\Lambda$. It can be set to zero.
The second statement
It automatically follows from the first statement (given by eq. $(1)$) if we use definition
$$
| \mathbf p \rangle = \sqrt{2E_{p}}a^{\dagger}(\mathbf p) | \rangle
$$
and eq. $(1)$, we'll get
$$
U(\Lambda )| \mathbf p \rangle = U(\Lambda )\left[\sqrt{2E_{p}} a^{\dagger}(\mathbf p)|\rangle\right] = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ]U(\Lambda ) | \rangle =
$$
$$
= \left| U(\Lambda ) | \rangle = | \rangle \right| = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ] | \rangle = \sqrt{2E_{\Lambda p}}\hat{a}^{\dagger}(\mathbf {\Lambda p })| \rangle \Rightarrow
$$
$$
U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_{p}}}\hat{a}^{\dagger}(\mathbf {\Lambda p }).
$$
Best Answer
I think it is correct. However, just as a pedantic remark, you should prove $U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)$ and $\sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}$ act on all the vectors in the same way, not just on the vacuum $|0\rangle$, you need a slight modification of your proof: $$\sqrt{2 E_{\mathbf{p}}}U(\Lambda)a^{s\dagger}_\mathbf{p}|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle=U(\Lambda)|\mathbf{p},s;\mathbf{p_1},s_1;\mathbf{p_2},s_2\cdots\rangle\\ =|\Lambda\mathbf{p},s;\Lambda(\mathbf{p_1},s_1);\Lambda(\mathbf{p_2},s_2)\cdots\rangle $$ The second equality is justified since we have assumed $s$ is parallel to the rotation or boost direction, and $s_1,s_2,\cdots$ might not, that is why I labeled the them differently as $\Lambda(\mathbf{p_1},s_1)$. To continue, we have $$=\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}|\Lambda(\mathbf{p_1},s_1);\Lambda(\mathbf{p_2},s_2)\cdots\rangle\\ =\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}U(\Lambda)|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle$$ Since the collection $\{|\mathbf{p_1},s_1;\mathbf{p_2},s_2;\cdots\rangle\}$ is a basis of the vector space, $$\sqrt{2 E_{\mathbf{p}}}U(\Lambda)a^{s\dagger}_\mathbf{p}=\sqrt{2 E_{\Lambda \mathbf{p}}} a^{s\dagger}_{\Lambda \mathbf{p}}U(\Lambda)$$ holds as an operator identity.