Recursive definitions of functions appear in virtually every area of mathematics, but I'm not aware of any area of mathematics that is devoted solely to their study. However, recursion and induction are studied in most detail in any area of mathematics that considers foundational issues more generally. The traditional foundation of mathematics is axiomatic set theory, and the most widely studied set theory is $\mathsf{ZFC}$. Here is what Paul Halmos has to say about recursion in his book Naive Set Theory (I have made a few modifications to the notation, but otherwise entirely preserved the quote):
Induction is often used not only to prove things but also to define things. Suppose, to be specific, that $f$ is a function from a set $X$ into the same set $X$, and suppose that $a$ is an element of $X$. It seems natural to try to define an infinite sequence $(u_n)_{n\in\mathbb N}$ of elements of $X$ (that is, a function $u$ from $\mathbb N$ to $X$) in some such way as this: write $u_0=a$, $u_1 = f(u_0)$, $u_2=f(u_1)$, and so on. If the would-be definer were pressed to explain the "and so on," he might lean on induction. What it all means, he might say, is that we define $u_0$ as $a$, and then, inductively, we define $u_{S(n)}$ as $f(u_n)$ for every $n$. This may sound plausible, but, as justification for an existential assertion, it is insufficient. The principle of mathematical induction does indeed prove, easily, that there can be at most one function satisfying all the stated conditions, but it does not establish the existence of such a function. What is needed is the following result.
Recursion theorem. If $a$ is an element of a set $X$, and if $f$ is a function from $X$ into $X$, then there exists a function $u$ from $\mathbb N$ into $X$ such that $u_0=a$ and such that $u_{S(n)} =f(u_n)$ for all $n$ in $\mathbb N$.
The proof of the recursion theorem can be found in Halmos's book, and in many other introductory books on mathematical logic and set theory. Once we have the recursion theorem in hand, we can now turn to your example. You want to prove that there is a sequence $u:\mathbb N\to\mathbb N$ such that $u_0=0$ and $u_{S(n)}=S(u_n)=u_n+1$. (I have rewritten your definition so as to avoid subtraction – subtraction in the context of the natural numbers is often awkward to deal with, as $a-b$ only makes sense when $a\ge b$.) In this particular case, we can recognise that $u:\mathbb N\to\mathbb N$ is simply the identity function on $\mathbb N$, and so it is easy to prove the existence of $u$: we can define the set $g=\{(x,y)\in\mathbb N\times\mathbb N:x=y\}$, and then prove by induction that $u=g$. However, what if $u$ had a more complicated definition? Perhaps we attempted to "define" $u$ by specifying a set $X$, an $a\in X$, and a function $\phi:\mathbb N\times X\to X$, before setting
\begin{align}
u_0&=a \\[3pt]
u_{S(n)}&=\phi(n,u_n) \, .
\end{align}
Here, it is not immediately clear how to apply the recursion theorem, as the definition of $u_{S(n)}$ appears to depend on both $n$ and $u_n$, not just $n$. However, with a bit more work, we can in fact prove a version of the recursion theorem that allows for these more general kinds of definitions. You can find the details in this post, for example.
