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
$$
\begin{align}
\frac{2}{E+m}\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=1-\frac{E^2-m^2}{(E+m)^2}+\left(1+\frac{E^2-m^2}{(E+m)^2}\right)\gamma^0-\frac{2}{E+m}p_j\gamma^j \\
2\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=(E+m)-(E-m)+(E+m+E-m)\gamma^0-2p_j\gamma^j \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+E\gamma^0-p_j\gamma^j\quad\quad\text{but}\quad p_\mu=(p_0,-\vec{p})=(E,-p_x,-p_y,-p_z) \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=m+p_\mu\gamma^\mu \\
\sum_{s=1}^2u_s(p)\overline{u}_s(p')&=\not p+m
\end{align}
$$
