Rings in which every non-unit is a zero divisor
A commutative ring $A$ has the property that every non-unit is a zero divisor if and only if the canonical map $A \to T(A)$ is an isomorphism, where $T(A)$ denotes the total ring of fractions of $A$. Also, every $T(A)$ has this property. Thus probably there will be no special terminology except "total rings of fractions".
Artinian rings provide examples: If $x \in A$, the chain $... \subseteq (x^2) \subseteq (x) \subseteq A$ is stationary, say $x^k = y x^{k+1}$ for some minimal $k \geq 0$. If $k=0$, $x$ is a unit. If $k \geq 1$, $x (x^k y - x^{k-1})=0$ and $x^{k-1} \neq y x^k$, i.e. $x$ is a zero divisor.
The class of total rings of fractions is closed under (infinite) products and directed unions. Is it the smallest such class containing the artinian rings?
Any (commutative unitary) ring of Krull dimension 0 has this property. This includes the class of Artinian rings.
[Edit] Proof: If $A$ has Krull dimension $0$, then any maximal $m$ ideal of $A$ is also a minimal prime ideal by the definition of Krull dimension. Applying Krull's theorem on the intersection of prime ideals to the localization $A_m$, we find that $mA_m$ is the nilradical of $A_m$. So for any $f\in m$, there exists a positive integer $n$ such that $f^n=0$ in $A_m$. So there exists $s\in A\setminus m$ such that $sf^n=0$ in $A$. We can choose $n$ smallest with respect to this property so that $sf^{n-1}\ne 0$. Therefore $f$ is a zero divisor. Now any non-unit element $f$ belong to some maximal ideal, it is a zero divisor.
Artining rings are zero-dimensional and semi-local. Boolean rings are reduced and zero-dimensional.
If $(A, m)$ is a local ring, then it has this property if and only if $\mathrm{depth}(A)=0$ (e.g. the example in Daniel Erman's comment). If $A$ is not necessarily local but all localizations $A_m$ at maximal ideals of $A$ have depth $0$, then $A$ has your property. But I don't think this is a necessary condition.
[Edit] Similarly one can construct local rings of depth 0 of any (even infinite) dimension. But I don't know whether there exists a ring of positive dimension with infinitely many maximal ideals and such that all its localizations at maximal ideals have depth 0.
Pace Chris Leary, the standard terminology is that a module all monomorphisms of which are automorphisms is said to be cohopfian (or co-Hopfian, if you're checking MathSciNet). A Dedekind-finite (a.k.a. directly finite) module usually means a module whose left invertible endomorphisms are also right invertible, equivalently, a module that is not isomorphic to any proper direct summand of itself.
T. Y. Lam, in his book Lectures on Modules and Rings, pp. 320–322, calls a noncommutative ring in which every regular element (i.e. neither right nor left zero-divisor) is a unit a classical ring, and he provides various examples. One that has not already been mentioned here is that any right (or left) self-injective ring is classical. Right self-injective rings need not have the property that every element that is merely not a left zero-divisor is a unit; interestingly, for right self-injective rings the latter condition is equivalent to the ring being Dedekind-finite (in the sense of the preceding paragraph), and also equivalent to the ring having stable range 1 (see Y. Suzuki, “On automorphisms of an injective module,” Proc. Japan Acad. 44 (1968), 120–124, and G. F. Birkenmeier, “On the cancellation of quasi-injective modules,” Comm. Algebra 4 (1976), no. 2, 101–109).