What is the purpose of a function being surjective?
Yes, we can arbitrarily only look at the range of functions, but this often misses the point. When we study a function, $f: A \rightarrow B$, often we're interested in the properties of $A$ and $B$ just as much as we are the properties of $f$. So, if we want to learn about $B$, and we know that we can do this somehow using surjective function $f$ from $A$ to $B$, just looking at the range of $f$ means we've given up looking at $B$, which is what we wanted to learn about in the first place.
Edit: Since this seems like we're just talking about very basic stuff, here's a very basic property. Let's say we want to know if $A$ and $B$ have the same cardinality. How do we know that? That there exists a bijective function $f: A \rightarrow B$. If we look at the range of $f$ instead of the codomain, we're no longer thinking about the cardinality of $B$, we're thinking about something else all together.
The other answers are good, but I'd like to add one thing. Suppose a function $f:\mathbb{R}\to\mathbb{R}$ is given. Is it possible to solve the equation $f(x)=b$, for some particular $b$?
If we know that $f$ is surjective, then we can be sure that a solution exists for any choice of $b$. If not, then we need to worry about whether $b$ is in the range of $f$ or not.
In linear algebra, this comes up a lot. The range of a linear function, given by a matrix $A$, so $f(x)=Ax$, is called the column space of $A$. Sometimes, the column space is the entire codomain, and sometimes it is a subspace. Whether or not a function like that is surjective becomes an interesting question, not only for solving equations, but for answering other questions about the structure of the function.
For example, what if the domain is $\mathbb{R}^4$, and the codomain is $\mathbb{R}^3$. Then what kind of subset of the domain solves the equation $f(x)=(0,0,0)$? If we know that $f$ is surjective, then we can answer that the set mapping to zero is a one-dimensional subspace. If $f$ is not surjective, then the set mapping to zero will have greater dimension.
$\newcommand{\Reals}{\mathbf{R}}\newcommand{Ratls}{\mathbf{Q}}$Mathematical problems often come in the form of,
"Some value $y$ depends deterministically on data $x$; is every prospective value $y$ an actual value?"
Two common formulations are:
Let $Y$ be a set of prospective values (specified in advance by the context of an external question), $X$ the set of allowable inputs, and $f:X \to Y$ a mapping representing the dependence $y = f(x)$.
The question above means Is $f$ surjective?
Let $Y \subset Z$ be a set of prospective values, $X$ the set of allowable inputs, and $f:X \to Z$ a mapping representing the dependence $y = f(x)$.
The question above means Is $Y \subset f(X)$?
Here's a selection of five examples, four of them kind of the same:
Is every real number the square of some real number?
That is, if $f:\Reals \to \Reals$ is defined by $f(x) = x^{2}$, is $f$ surjective? (Answer: No. For instance, $-1$ is not in the image.)
Is every non-negative real number the square of some real number?
That is, if $f:\Reals \to \Reals$ is defined by $f(x) = x^{2}$, is $[0,\infty)$ contained in the image of $f$? (Answer: Yes, though proving this "existence of real square roots" requires non-trivial use of the completeness axiom for the real numbers, even though the result is usually introduced into the curriculum many years prior to a careful analysis course.)
Is every positive rational number the square of some rational number?
That is, if $f:\Ratls \to \Ratls$ is defined by $f(x) = x^{2}$, is $[0,\infty) \cap \Ratls$ contained in the image of $f$? (Answer: No. For instance, $2$ is not in the image.)
If $y:\Reals \to \Reals$ is a continuous function, does there exist a differentiable function $x:\Reals \to \Reals$ such that $x' = y$?
That is, if $X$ is the set of differentiable, real-valued functions on $\Reals$, and $Y$ is the space of continuous functions, and $Z$ the space of all functions, and if $f(x) = x'$, is $Y$ contained in the image of $X$? (Yes: One of the fundamental theorems of calculus guarantees that every continuous function on $\Reals$ is the derivative of some differentiable function.)
Let $(M, g_{0})$ be a compact Kähler manifold. If $\rho$ is a smooth $(1, 1)$-form in the cohomology class $2\pi\, c_{1}(M)$, does there exist a Kähler metric $g$ whose Kähler form is cohomologous to the Kähler form of $g_{0}$ and whose Ricci form is $\rho$?
Analogously to the preceding example, you can imagine that there is a partial differential equation of the abstract form $\rho = f(g)$, and the question amounts to surjectivity of the Ricci curvature operator $f$. The answer turns out to be "yes"; largely for this resolution of the Calabi conjecture, S. T. Yau was awarded the Fields Medal in 1982.
The take-away is, not only is surjectivity interesting, but proving that a specific mapping is surjective can constitute a major work in a distinguished mathematical career.