First of all, there are many rings without units that are very interesting. A prominent example is the $C^*$-algebra $C_0(X)$, that is $C_0(X)=\left\{f\in C(X)\mid \mbox{ $f$ vanishes at infinity}\right\}$. Here $X$ is a Hausdorff topological space. In fact, $C_0(X)$ has a unit if and only if $X$ is compact. These spaces are very important since any commutative $C^*$-algebra is of this form.
Now given a ring $R$ without unit, one can always embed $R$ into a unital ring of the same characteristic (which can still be defined for non-unital rings). So one can use the usual ring theory on this unitization. (A unitization is not unique, so the usefulness depends on what you try to do).
Obviously if you work in a non-unital setting, ring theory becomes more difficult. Normally if $R$ and $S$ are rings and $f:R\rightarrow S$ is a ring morphism, one requires that $f(1_R)=1_S$, in a non-unital setting we no longer have this and thus ring morphisms are more difficult.
Your question is very general, and there are many things that can go wrong. For example if $R$ is a ring with unit, then any ideal in the matrix ring $M_{n\times n}(R)$ is of the form $M_{n\times n}(A)$. This is no longer true if $R$ has no unit. Hence the theory of matrices over non-unital rings is quite different.