Call a character $\chi$ (left) generating if every character is of the form $\psi(a)=\chi(ab)$ for some $b\in R$. It turns out that, when $R$ is finite, a character is left generating if and only if it is right generating (every character has the form $\psi(a) = \chi(ba) $ for some $b\in R$) when $R$ is finite.
The answer to your question is as follows:
A finite ring $R$ has a generating character if and only if $R$ is a finite Frobenius ringFrobenius ring.
For a proof, see, for example:
Thomas Honold, MR 1831096 Characterization of finite Frobenius rings, Arch. Math. (Basel) 76 (2001), no. 6, 406--415.
When the ring is finite and commutative, this is equivalent to $R$ having a multiplicity-free socle, or $R$ being a direct product of finite local rings with simple socle. (see T. Y. Lam, MR 1653294 Lectures on modules and rings , Thm. 15.27, and more generally §§15, 16 for more characterizations and examples and non-examples.)
Your question is quite natural and appears in other contexts. For example, it appears naturally when studying Gauss sums over arbitrary finite rings, as in a series of papers by Lamprecht (cited in Honold's paper), and in:
Fernando Szechtman, MR 1916077 Quadratic Gauss sums over finite commutative rings, J. Number Theory 95 (2002), no. 1, 1--13.
This property of finite Frobenius rings is also relevant to coding theory, see
Jay A. Wood, MR 1738408 Duality for modules over finite rings and applications to coding theory, Amer. J. Math. 121 (1999), no. 3, 555--575.