What is subtraction? Let $a, b \in \mathbf{R}$. It is an elementary fact that addition is a commutative binary operation on the reals, that is, $a + b \in \mathbf{R}$ and $a + b = b + a$. With the exception of commuativity, the same goes for subtraction. It is indeed clear that $a -b \ne b - a$ for some $a, b$. However, consider the intuitive notion of subtraction where one simply performs addition with the additive inverse of $b$:$$a + (-b) = (-b) + a$$
If one uses the intuition of "add the negative of the number" as a means of subtraction, you are simply using addition, a commutative operation. Of course, it is still absurd to ask the question "Is subtraction commutative?", seeing as we no longer consider subtraction as a binary operation, but addition on the additive inverse. 
While the notion of subtraction as a binary operation is the one that is most widely used in elementary schooling (that is, a third grader will likely be clueless about binary operations, but he won't argue that $3 -4 = 4 - 3$), the second allows for the construction of algebraic structures such as the abelian group $\langle \mathbf{Z}, + \rangle$. I note that in the few introductory courses I have participated in, and in particular that of Abstract Algebra, this second notion of subtraction is used more widely.
Is subtraction (for simplicity, on the reals) the binary operation $-$, or the additive operation of $a + (-b)$. In particular, does subtraction as a binary operation serve a(n important) purpose in the research of algebraic structures?
 A: to define $\mathbb{Z}$ from $\mathbb{N}$, you take equivalence classes of pairs natural numbers, with $(a,b)=(c,d)$ if $a+d=b+c$ (intuitively, $(a,b)=a-b$) and define addition coordinate-wise.  so, we are formally adjoining additive inverses, $(a,b)+(b,a)=(0,0)$.
similarly, to construct $\mathbb{Q}$ from $\mathbb{Z}$, take equivalence classes of pairs of integers $(a,b)=(c,d)$ if $ad=bc$ (intuitively $(a,b)=a/b$) and define multiplication component-wise, addition via $(a,b)+(c,d)=(ad+bc,bd)$.
you can ask the same questions about division as you do about subtraction: it isn't commutative, or associative; it too is a unary operation $x\to x^{-1}$.  we are formally adjoining multiplicative inverses.
A: In response to your comment "we no longer consider subtraction as a binary operation, but addition on the additive inverse":
A binary operation is a rule which takes two elements of your set, taken in a specified order, and assigns to them a new element. As such a general binary operation can be defined in whatever way you like. For example, I could define an operation $a \mathrel{\substack{\sim\\*\\ \sim}} b$ in natural numbers to mean "take $a+1$ and concatenate it with the sum of the digits of $b+5$". It's a perfectly well-defined operation. It doesn't do anything useful, is not commutative or associative or anything else nice, but it is a binary operation.
Even if $a-b$ is defined to be $a + (-b)$, then it is still a binary operation. It has merely been constructed from another binary operation, but that doesn't make it any less of an operation in its own right.
A question such as "Is subtraction commutative?" is perfectly reasonable even if it is $a + (-b)$. It is simply asking the question: "Is $a + (-b)$ the same as $b + (-a)$ for all $a$ and $b$?".

In response to "Is subtraction (for simplicity, on the reals) the binary operation −, or the additive operation of $a+(−b)$?":
The above argument indicates that it is in fact both. Technically the "binary operation -" doesn't mean anything until it is defined, and depending on your situation it is defined differently. Some possible definitions are:


*

*For $\mathbb{N}$, $a-b$ could be defined as the result of starting with $a$ and performing the inverse of the successor function $b$ times.

*For $\mathbb{Z}$, $a-b$ could be defined as the solution $x$ to $a = b+x$. This works in $\mathbb{N}$ too, but the solution is undefined for many combinations.

*For $\mathbb{R}$, you can first define $-b$ as the additive inverse of $b$ (ie the unique number such that $b + (-b) =0$), and then you can define $a-b$ as $a + (-b)$.

*For a general additive group, the above defintion is the onlyone that usually makes sense, and in places where there is an alternative definition of - it is designed to match perfectly.



As to whether it is important, well I'm not sure, but it is useful as an example of a binary operation which is neither commutative nor associative. Also, one of the definitions of a subgroup of an additive group is that it is a nonempty subset that is closed under subtraction (where subtraction is defined via addition).
A: I may be in the wrong here, but I think of substraction (well, from an abstract algebra point of view...) as :


*

*first you enrich the monoid $(\mathbb{N},+)$ with symbolic elements $-n$ to get a group

*now you have a group $(\mathbb{Z},+)$, and $a-b = a + (-b)$ by construction/definition.
Of course this carries over to $\mathbb{R}$.
If you want to think about substraction of elements of $\mathbb{N}$ "while only knowing about $\mathbb{N}$", then I think it's reasonnable to think of it as the solution of 
$$a=b+x$$ 
