I think a major reason is because Lie algebras don't have an identity, but I'm not really sure.

The reason is simple: There are many nonunital 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 continous complexvalued 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 antiequivalence between unital commutative $C^\ast$algebras and compact spaces, and also between (possibly nonunital) 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 antiequivalence between unital boolean rings and compact totally disconnected spaces, and also between boolean rings and locally compact totally disconnected spaces. OnepointCompactification 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 nonunital rings. If $A$ is a ring (possibly nonunital), 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 nonunital 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? Becuase 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 infinitedimensional, so this precludes one of the most powerful things that a representation theorist has in his arsenalnamely 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 locallyconstant complexvalued 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 locallyconstant 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 algebrathe socalled 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 algebrait 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 complexvalued functions on $[0,\infty)$ with the "multiplication" operation of convolution... $$ f * g (x) = \int_0^x f(t) g(xt)\,dt $$ 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 


A lowlevel answer, but I found it pretty surprising: Dimension shifting for Hochschild cohomology is easier to prove for nonunital rings than for unital rings. Let me explain these notions: Let $A$ be a (not necessarily unital) $k$algebra (with $k$ a commutative ring), and $P$ an $\left(A,A\right)$bimodule. We denote by $C^n\left(A,P\right)$ the (additive) $k$module of all $k$linear homomorphisms $A^{\otimes n}\to P$. We define the differential $\delta:C^n\left(A,P\right)\to C^{n+1}\left(A,P\right)$ by $\left(\delta f\right)\left(a_1\otimes a_2\otimes ...\otimes a_{n+1}\right)$ $= a_1 f\left(a_2\otimes a_3\otimes ...\otimes a_{n+1}\right) + \sum\limits_{i=1}^n \left(1\right)^i f\left(a_1\otimes a_2\otimes ...\otimes a_{i1} \otimes a_i a_{i+1} \otimes a_{i+2} \otimes a_{i+3} \otimes ... \otimes a_{n+1}\right)$ $ + \left(1\right)^{n+1} f\left(a_1\otimes a_2\otimes ...\otimes a_n\right) a_{n+1}$. This satisfies $\delta²=0$, so we get a cohomology $k$module $H^n\left(A,P\right)$, which is called the $k$th Hochschild cohomology of $A$ and $P$. Dimensionshifting now states that $H^{m+1}\left(A,P\right) = H^m\left(A,Q\right)$ for any $m\geq 1$, where the $\left(A,A\right)$bimodule $Q$ is the $k$vector space $C^1\left(A,P\right)=\mathrm{Hom}_k\left(A,P\right)$ with $\left(A,A\right)$bimodule structure defined by $\left(a*f\right)\left(b\right)=a\cdot f\left(b\right)$ for any $a\in A$, $f\in Q$, $b\in A$; $\left(f*a\right)\left(b\right)=f\left(ab\right)f\left(a\right)b$ for any $a\in A$, $f\in Q$, $b\in A$. Now, if you try to do this all for rings $A$ with unity and for unital $\left(A,A\right)$bimodules $P$ (id est, the unity of $A$ acts as identity from both sides on $P$), you are in for a bad surprise: Even if $P$ is a unital $A$module, $Q$ isn't necessarily. It's the right $A$action which causes the troubles. What you can do instead is replacing $Q$ by the subset of $Q$ formed by all those $f\in Q$ which satisfy $f\left(1\right)=0$. But now proving $H^{m+1}\left(A,P\right) = H^m\left(A,Q\right)$ isn't as easy anymore, as we have to show that cohomology of normalized cochains is the same as cohomology of cochains (this amounts to finding a chain homotopy, something which is implicit in Hochschild's Annals 1946 paper). 


Perhaps you will find the following remarks of interest, excerpted from the preface of Gardner and Wiegandt: Radical Theory of Rings, 2004. 


To answer a slightly generalized question, there are nonunital ring maps between unital rings that come up naturally. If $e \in A$ is an idempotent, then the ring of elements of the form $eAe$ inherits its additive and multiplicative structure from $A$, but its identity element is $e$ and not $1_A$. For example, if $k$ is a commutative ring and $m < n$ then the map $M_m(k) \to M_n(k)$ given by "padding by 0's" is a natural nonunital map of unital algebras. Under certain circumstances the rings $A$ and $eAe$ are Morita equivalent, so this type of situation can be useful in representation theory. 


For work related to radicals of rings, the Köthe Conjecture, etc., it's very useful to consider "rngs" (Louis Rowen's term for rings without identity). 

