A ring $A$ is said to have the ITI property with respect to an ideal $\mathfrak{a}$ if $\mathfrak{a}$-torsion submodules of injective $A$-modules are injective.

Noetherian rings have ITI with respect to every ideal, but a ring with ITI with respect to every ideal is not necessarily noetherian (e.g. absolutely flat rings have ITI). However, there are rings without ITI, even with respect to some principal ideals.

If $A$ has the ITI property with respect to (the ideal generated by) $f$ then the canonical morphism $I\rightarrow I_f$ is an epimorphism. Hence, noetheriannes is not necessary for this property to hold. But it is unknown to me (and I think also in general unknown) whether noetheriannes can be omitted.

See also this question and its answer.

ADDENDUM: Besides the ITI property, there are other hypotheses on the ring $A$ that imply that the canonical morphism $I\rightarrow I_f$ is an epimorphism for every injective $A$-module $I$. Namely, this is true if $A$ is hereditary, or if every zerodivisor of $A$ is nilpotent, hence in particular if $A$ is integral.