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 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.
