The definition of product in Lang's Algebra (page 58) is this:
Let $(P,f,g)$ be a product of $A$ and $B$. Let be $C=A$ and $\varphi=id_A$. Then, by definition, there is a (unique) $h:A\to P$ morphism so, that $id_A=f\circ h$, that is, $h$ is a right inverse of $f$, that is, $f$ is a retraction. The similar holds for $g$.
Is this proof correct? I'm pretty sure that yes, but I'm a bit surprised that I didn't find this statement anywhere. That's why I need a confirmation.
On
That seems ok, as long as there are arrows $A \to B$, because the existence of $h$ will be guaranteed provided that you give morphims $C \to A$ and $C \to B$, in which case $\phi = fh$ and $\psi = gh$ ought to exist.
If you have any arrow $a : A \to B$ then $(id_A,a)$ factors through $P$ via some unique $h$ as you have said.
As drhab has said in his answer, there are rather elementary examples in which this fails already. I know little category theory, but maybe more 'interesting' examples can be fabricated out of well known objects with no arrows between them.
Almost.
It only works if some arrow $A\to B$ indeed exists.
If we are e.g. working in the category of sets with $B=\varnothing$ and $A\neq\varnothing$ then this is not the case and also $P=A\times B=\varnothing$.
In that case $f:P=\varnothing\to A\neq\varnothing$ has no right-inverse.