We were discussing about the possibility of having an algebra over a field which is associative but has not the unity. Does it exist? It has been proposed as a counterexample the set of even numbers. We believe it is wrong because it can't be a vector space. If an associative but not unital algebra exists, how can we write its associative property? We usually write it as a commutative diagram or in terms of maps composition. In these terms the identity seams unavoidable. i.e. $$ m\circ\left(m\otimes id\right) = m\circ\left(id\otimes m\right)\qquad associativity$$ with $m : A\otimes A\rightarrow A$

The set of even numbers is a nonunital ring, in particular it has an associative multiplication, but you are right that it isn't an algebra over a field. For an example of a nonunital algebra, consider the continuous functions $\mathbb{R} \to \mathbb{R}$ which vanish at a particular point, under pointwise product. More generally, any proper twosided ideal of an algebra is a nonunital algebra, just as any proper twosided ideal of a ring is a nonunital ring. You may think that these examples are "unnatural," so here is a "natural" one: the algebra of compactly supported continuous functions $\mathbb{R} \to \mathbb{R}$ under convolution. For an example of a nonassociative algebra, take, for example, the octonions. Your confusion arises because of the following issue: given a vector space $V$ and a bilinear operation $V \times V \to V$ we can associate to any $a \in V$ the linear operator $L_a$ which is left multiplication by $a$, and these linear operators form an associative algebra. However, composition of the operators $L_a$ need not be the same as $\times$: associativity is equivalent to the statement that $L_a L_b = L_{a \times b}$, which need not be the case in general. I think this is the source of your confusion. 

