Proof:

If $P^{\prime}$ with morphisms $f^{\prime}_1:A_1\longrightarrow P^{\prime}$ and $f^{\prime}_2:A_2\longrightarrow P^{\prime}$ is another candidate we may apply the universal property to get a map $\phi:P\longrightarrow P^{\prime}$ such that

$$\phi\circ f_1 = f^{\prime}_1,\quad \phi\circ f_2 = f^{\prime}_2.$$

Reciprocally since $P^{\prime}$ is a push out, there is a map $\psi:P^{\prime}\longrightarrow P$ such that

$$\psi\circ f^{\prime}_1 = f_1,\quad \psi\circ f^{\prime}_2 = f_2.$$

Combining we see that $(\psi\circ\phi)\circ f_1 = f_1$ and $(\psi\circ\phi)\circ f_2 = f_2$. We see that both $\psi\circ\phi$ and id$_{P}$ satisfy the universal property with $E = P$, $g_1 = f_1$ and $g_2 = f_2$. The uniqueness clause in the definition gives $\psi\circ\phi =$   id$_{P}$. Likewise we get $\phi\circ\psi =$   id$_{P^{\prime}}$ and the proof is complete.