What are the reasons for considering rings without identity?
The reason is simple: There are many non-unital rings which appear quite naturally.
If $X$ is a locally compact space (in the following every space is assumed to be Hausdorff), then $C_0(X)$, the ring of continuous complex-valued functions on $X$ vanishing at infinity, is a $C^\ast$-algebra which is unital if and only if $X$ is compact. If $X = \mathbb{N}$, this is just the ring of sequences converging to $0$. Gelfand duality yields an anti-equivalence between unital commutative $C^\ast$-algebras and compact spaces, and also between (possibly non-unital) commutative $C^*$-algebras (with "proper" homomorphisms) and locally compact spaces (with proper maps). In a very similar spirit ($\mathbb{C}$ is replaced by $\mathbb{F}_2$), there is an anti-equivalence between unital boolean rings and compact totally disconnected spaces, and also between Boolean rings and locally compact totally disconnected spaces. One-point-Compactification on the topological side corresponds here to the unitalization on the algebraic side. Perhaps we have the following conclusion: As locally compact spaces appear very naturally in mathematics (e.g. manifolds), the same is true for non-unital rings.
If $A$ is a ring (possibly non-unital), its unitalization is defined to be the universal arrow from $A$ to the forgetful functor from unital rings to rings. An explicit construction is given by $\tilde{A} = A \oplus \mathbb{Z}$ as abelian group with the obvious multiplication so that $A \subseteq \tilde{A}$ is an ideal and $1 \in \mathbb{Z}$ is the identity. Because of the universal property, the module categories of $A$ and $\tilde{A}$ are isomorphic. Thus many results for unital rings take over to non-unital rings.
Every ideal of a ring can be considered as a ring. Important examples also come from functional analysis, such as the ideal of compact operators on a Hilbert space.
What is the reason for considering any algebraic structure? Because it comes up naturally when trying to do other things!
Here's a concrete example. In the Langlands programme one of the main local conjectures is relating representations of a (connected reductive) $p$-adic group to representations of a (group related to a) Galois group. Now most of the interesting representations of the $p$-adic group are infinite-dimensional, so this precludes one of the most powerful things that a representation theorist has in his arsenal---namely the possibility of taking traces. But in fact this can be fixed up very nicely! There is an analogue of the "group ring" of our $p$-adic group, namely the space of locally-constant complex-valued functions on the group with compact support. This space interits an addition (obvious) and a multiplication (convolution: the group has a natural measure on it, namely the Haar measure). So it's an algebra. Furthermore it is easily checked to have no identity element (the "delta function" isn't a locally-constant function!). However it's also not hard to check that there's an equivalence of categories between (certain) representations of the $p$-adic group that one is interested in, and (certain) representations of this algebra---the so-called Hecke algebra. Furthermore elements of the Hecke algebra act via maps with finite image, and so have traces! This is a big win. One can prove linear independence of characters etc etc, and get the powerful techniques back. But no way can the identity map be in this Hecke algebra---it certainly doesn't have finite image in general, and hence no trace.
Representations of the Hecke algebra are absolutely crucial in many works on this part of the Langlands correspondence, but they have no identity element. So there is one reason, in my area, at least.
Here is a favorite example. (See also Martin's answer.) Consider $C[0,\infty)$, the continuous complex-valued functions on $[0,\infty)$ with the "multiplication" operation of convolution... $$ f * g (x) = \int_0^x f(t) g(x-t)\,dt $$ It is a ring. Without unit. Even an integral domain. Mikusinski[*] said, take the field of fractions. Great. A simple introduction to generalized functions. Now if the student had studied algebra from some perverse textbook that constructed the field of fractions only in the unital case, what is the student to do? Go back to the textbook and check that it works without unit? A good exercise for that student, I guess.
[*] Jan Mikusinski, OPERATIONAL CALCULUS, 1959