Suppose $f: A →B$ and $s: B → A$ and $f∘s=1_B$ and for every object T and any pair of maps $t_1$ and $t_2$ such that $t_1: T → A$ and $t_2 : T → A$, it is true that if $f∘t_1=f∘t_2$ then $t_1 = t_2$. Note that A, B and C are objects in a general category.
Question: If above conditions hold, can it be proved that $f$ is an isomorphism using only basic general axioms of a category?
This answer is based upon @LordSharktheUnknown 's hint.
Consider the map $f \circ (s\circ f):A\rightarrow B$.
Using associative property of composition of maps, $$f \circ (s\circ f)=(f\circ s)\circ f.$$
Now, since $s$ is a section for $f$, $$(f\circ s)\circ f=1_B\circ f=f.$$
Now we have shown that, $$f\circ (s\circ f)=f.$$
But also using identity law, $$f\circ 1_A=f.$$
This was what @LordSharktheUnknown's hint expressed as $f\circ (s\circ f)=f\circ 1_A$.
Since $f$ is a monomorphism, by definition, $$[f\circ (s\circ f)=f\circ 1_A]\Rightarrow [s\circ f = 1_A].$$
Now, we have shown that,
$$s\circ f = 1_A (New)\ and \ f\circ s = 1_B \ (Given).$$ Therefore, $f$ is an isomorphism, and $s$ is its inverse.(Q.E.D.)
Notice, similarly, given retraction $r$ for a $f$ and $f$ being epimorphism, it may be, in which I have high degree of confidence, showed that $f$ is isomorphism.