Ok, so here I found an answer. I'm gonna start rewriting the last expression of the question:
$$
\begin{align}
\sum_{s=1}^2u_s(p)\overline{u}_s(p') &= \sqrt{(E+m)(E'+m')}
\begin{pmatrix}
1 & -\frac{\vec{\sigma}\cdot\vec{p'}}{E'+m'} \\
\frac{\vec{\sigma}\cdot\vec{p}}{E+m} & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{(E+m)(E'+m')} \\
\end{pmatrix} \\
\frac{1}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p') &= \Omega
\end{align}
$$
where I have defined for convenience
$$
s:=E+m \quad\quad s':=E'+m' \quad\quad \Omega:=
\begin{pmatrix}
1 & -\frac{\vec{\sigma}\cdot\vec{p'}}{s'} \\
\frac{\vec{\sigma}\cdot\vec{p}}{s} & -\frac{(\vec{\sigma}\cdot\vec{p})(\vec{\sigma}\cdot\vec{p'})}{ss'} \\
\end{pmatrix}
$$
Now, I know that $\Omega\in M(4,\mathbb{C})$ and a basis for this linear space is given by the set
$$
\{1,\gamma^\mu,\gamma^5,\gamma^\mu\gamma^5\}_{\mu=0}^3\cup\{\sigma^{\mu\nu}\}_{\mu,\nu=0\; (\mu<\nu)}^3
$$
(where $\sigma^{\mu\nu}:=i/2[\gamma^\mu,\gamma^\nu]$) so that $\Omega$ can be written as
$$
\Omega=a+b_\mu\gamma^\mu+c\gamma^5+d_\mu\gamma^\mu\gamma^5+e_{\mu\nu}\sigma^{\mu\nu}
$$
and where the coefficients are given by
- $a=\frac{1}{4}tr(\Omega)$
- $b_\mu=\frac{1}{4}tr(\Omega\gamma_\mu)$
- $c=\frac{1}{4}tr(\Omega\gamma_5)$
- $d_\mu=\frac{1}{4}tr(\Omega\gamma_5\gamma_\mu)$
- $e_{\mu\nu}=\frac{1}{8}tr(\Omega\sigma_{\mu\nu})$
With this, it follows that:
$$
\begin{align}
a&=\frac{1}{2}\left(1-\frac{\vec{p}\cdot\vec{p}'}{ss'}\right) \\
b_0&=\frac{1}{2}\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right) \\
b_j&=-\frac{1}{2}\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\quad j=1,2,3 \\
c&=0 \\
d_0&=0 \\
d_j&=\frac{i}{2ss'}(\vec{p}\times\vec{p}')_j\quad j=1,2,3 \\
e_{0j}&=-\frac{i}{4}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\quad j=1,2,3 \\
e_{jk}&=-\frac{i\varepsilon_{jkl}}{4ss'}(\vec{p}\times\vec{p}')_l\quad j=1,2,3\;\;\text{and}\;\; j<k
\end{align}
$$
Therefore,
$$
\frac{2}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p')=1-\frac{\vec{p}\cdot\vec{p}'}{ss'}+\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right)\gamma^0-\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\gamma^j+\frac{i}{ss'}(\vec{p}\times\vec{p}')_j\gamma^j\gamma^5-\frac{i}{2}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\sigma^{0j}-\frac{i\varepsilon_{jkl}}{2ss'}(\vec{p}\times\vec{p}')_l\sigma^{jl}
$$
or, using the following identities:
$$
\sigma^{0j}=i\gamma^0\gamma^j\quad\quad \varepsilon_{jkl}\varepsilon^{jkn}=2\delta_l^n \quad\quad\sigma^{jk}=\begin{pmatrix}
\varepsilon^{jkn}\sigma_n & 0 \\
0 & \varepsilon^{jkn}\sigma_n \\
\end{pmatrix}
=\varepsilon^{jkn}\sigma_nI_4
$$
then
$$
\frac{2}{\sqrt{ss'}}\sum_{s=1}^2u_s(p)\overline{u}_s(p')=1-\frac{\vec{p}\cdot\vec{p}'}{ss'}+\left(1+\frac{\vec{p}\cdot\vec{p}'}{ss'}\right)\gamma^0-\left(\frac{p_j}{s}+\frac{p'_j}{s'}\right)\gamma^j+\frac{i}{ss'}(\vec{p}\times\vec{p}')_j\gamma^j\gamma^5+\frac{1}{2}\left(\frac{p_j}{s}-\frac{p'_j}{s'}\right)\gamma^0\gamma^j-\frac{i}{ss'}(\vec{p}\times\vec{p}')\cdot\vec{\sigma}I_4
$$
As a particular case, if we assume now that both spinors correspond to the same 1/2-spin particle, then $p=p'$, $ss'=(E+m)^2$, $\vec{p}\cdot\vec{p}'=E^2-m^2$ and $\vec{p}\times\vec{p}'=0$ and the above expression reduces to
$$
\begin{align}
\frac{2}{E+m}\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=1-\frac{E^2-m^2}{(E+m)^2}+\left(1+\frac{E^2-m^2}{(E+m)^2}\right)\gamma^0-\frac{2}{E+m}p_j\gamma^j \\
2\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=(E+m)-(E-m)+(E+m+E-m)\gamma^0-2p_j\gamma^j \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+E\gamma^0-p_j\gamma^j\quad\quad\text{but}\quad p_\mu=(p_0,-\vec{p})=(E,-p_x,-p_y,-p_z) \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+p_\mu\gamma^\mu \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=\not p+m
\end{align}
$$
A 360° rotation of a spin 1/2 wavefunction does indeed produce a '-' sign. You can find more details in the chapter on angular momentum in Sakurai's Modern Quantum Mechanics.
Of course, this minus sign does not affect any observables, because we are calculating probabilities or expectation values, where the - signs on the bra and ket cancel each other out. However, it can be experimentally verified through interferometry experiments - We use a beam splitter on a monoenergetic beam of neutrons to create two paths. Introduce a phase change (=rotation of the ket, for example, using a magnetic field) in one path, and see whether the max/min interference condition is repeated for a phase change corresponding to a 360° rotation or 720°. Turns out the quantum mechanical prediction is right!
Best Answer
Yes. It can be seen easily by using a very utilized change in QM. You can see that $$ \vec{\sigma}_1 \vec{\sigma}_2 = \frac{1}{2} ( (\vec{\sigma}_1+\vec{\sigma}_2)^2 - \vec{\sigma}_1^2 -\vec{\sigma}_2^2 )$$
N.B. here we use the fact that $\vec{\sigma}_1 \vec{\sigma}_2 = \vec{\sigma}_2 \vec{\sigma}_1$ since they operate on different Hilbert spaces.
Now, from Pauli matrices algebra $\vec{\sigma}^2 = 3 $ therefore $$ \vec{\sigma}_1 \vec{\sigma}_2 = -3 +\frac{1}{2} (\vec{\sigma}_1 +\vec{\sigma}_2)^2 = -3 +2 (\vec{S}_1+\vec{S}_2)^2 $$
since $\vec{S} = \frac{1}{2} \vec{\sigma}$.
So the spin-exchange can be written as $$ \frac{1}{2}(1+ \vec{\sigma}_1 \vec{\sigma}_2 ) = -1 + (\vec{S}_1 + \vec{S}_2)^2 $$ You should know from spin addition algebra that there are a singlet state and three triplet states that are eigenstates of $(\vec{S}_1 + \vec{S}_2)^2 $ with eigenvalues 0 for the singlet and 2 for the triplets, therefore there are four linear independent states for which the this operator has a diagonalan representation as: $(\vec{S}_1 + \vec{S}_2)^2 = diag(0,2,2,2)$
and therefore, on the same basis, the spin exchange operator have diagonal representation as:
$$ \frac{1}{2}(1+ \vec{\sigma}_1 \vec{\sigma}_2 ) = diag(-1,1,1,1 ) $$
The square of this matrix is clearly the identity as you wanted to prove.