Can $f(\infty)$ be defined if the sequence $f(n)$ is divergent?
Infinity isn't a value, so saying $f(\infty)$ and $f(\infty + 1)$ is meaningless mathematically.
However,
$$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} f(x+1) = L,$$
but only if $L$ exists. This formalizes the argument you made.
It's a formal procedure to find the fixed point. Even if $|a|>1$, $\frac{b}{1-a}$ is still a fixed point of $f$. The difference is that if $|a| < 1$, $f$ converges to $\frac{b}{1-a}$ regardless of what $f(1)$ is, but if $|a|\geq1$ you only stay at the fixed point if that's where you started, while other points get moved off to infinity.
In fact, you'll notice that, formally, $\infty$ is also a fixed point. What this is saying is that $|a| < 1$ means that other points get sent to $\frac{b}{1-a}$, but if $|a|\geq1$ then other points get sent to $\infty$.
In short, the formal calculation does get a result that's meaningful in some sense, but not according to the usual definitions of limits. I'd draw a parallel to some interesting results you get from manipulating divergent series.
Long story short, your argument assumes that the function is convergent, i.e., that $f(\infty)$ is both finite and constant. But what if $f(\infty)$ is either infinite, or a set of values, instead of a single fixed point; i.e., $\sin(\infty)$ would be an entire interval, namely $[-1,+1]$, as opposed to a single fixed value.
As other answers have pointed out, $f(\infty)$ is defined if and only if you define it. Furthermore, you can define it in such a way that your formal manipulation is in fact a valid argument. So to me the question is not so much "is this valid?" but "is this useful?" Specifically, I'd like a definition such that:
Your formal argument applies and correctly computes the limit.
The definition applies to many different sorts of functions, and gives interesting answers for all of them.
I think such a definition exists. Moreover, there is a general principle for making interesting definitions of this type. This principle is analytic continuation.
The idea is simple: Express the limit or sum you want to study as a sum of a power series at a particular point. When the power series converges, you recover the usual answer. When it diverges, you try to sum the power series at a different point, forming a complex analytic function, and continue that analytic function around the singularity to your chosen point.
For instance, to the sequence $f(n)$ we could associate the power series $\sum_{n=1}^\infty f(n) x^n$. Observe that when $f(n)$ has a limit $f(\infty)$, as $x$ approaches $1$, this sum is $f(\infty)/(1-x)+ o(1/(1-x))$. So let's set:
$$g(x) = \sum_{n=1}^\infty f(n) x^n$$
$$ h(x) = (1-x) g(x)$$
and define $f(\infty)= h(1)$.
Now when the sum does not converge near $1$, but still converges near $0$, we can analytically continue $g(x)$. For the function you wrote down, it will always converge in some disc around $0$. Moreover, we may calculate from the recurrence relation:
$$ g(x) = \sum_{n=1}^\infty f(n) x^n = f(0) + \sum_{n=1}^{\infty} f(n+1) x^n = f(0) + ax g(x) + b \frac{x}{1-x} $$
$$g(x) = \frac{f(0) + b \frac{x}{1-x} }{1-ax}$$
$$h(x) = \frac{ f(0) (1-x) + bx }{1-ax}$$
These formulas describe $g$ and $h$ in the disc around $0$. Because these functions are analytic, they must also describe the analytic continuation of $g$ and $h$ wherever they are defined. Hence it is mathematically valid to compute:
$$h(1) = \frac{ f(0)(1-1) + b}{1-a} = \frac{b}{1-a}$$
This technique can be used to justify many other strange identities by appropriately defining the terms, such as the well-known "formula":
$$1+ 2+ 3+ 4 + \dots = \frac{-1}{12}$$
But, crucially, the analytic continuation of these sorts of sums is also crucial in many sorts of mathematical investigation - most famously the study of the Riemann zeta function.
It is also used in physics under the names "regularization" and "renormalization", but I don't know much about that.
When you are working with a "silly" formal manipulation that seems to produce an interesting result, don't ask yourself "is this rigorous?" but instead "how can I make this rigorous with appropriate definitions?" This mode of investigation has led to important insights in the past, and will lead to many more in the future. Maybe someday all symbolic manipulations used in physics will be given firm rigorous footing in their own mathematical theories that will have applications far beyond their original purpose.
You ask the question, is $f(\infty)$ defined? Well $f(x)$ usually means take the number in the codomain of $f$ associated with the value $x$ in the domain of $f$. Since you defined $f$ to have a domain of $\mathbb N$, and $\infty\not\in\mathbb N$, this doesn't make sense. There are however several adhoc ways we can fix this, and look at the bottom of this answer, for how to solve this problem properly.
First some of the adhoc ways of solving this:
You obviously mean something different than applying the function with $\infty$. One way to define this faulty notation $f(\infty)$, could be using limits as you did:
$$f(\infty):=\lim_{n\to\infty}f(n)$$
This interpretation makes reasonable sense, and results in different values for different $a$ and $b$, for example $a\ge1$ results in $f(\infty)=\pm\infty$ depening on $b$, and $|a|<1$ results in $f(\infty)=b/(1-a)$ and so on, I don't think finding these limits are at the heart of the problem.
Another way of interpreting this is saying that $\infty+1=\infty$ and plopping $\infty$ in the original definition, while ignoring that it's not defined for $\infty$. Then you get: $$ \begin{align} f(\infty+1) &= a f(\infty) + b\\ f(\infty) &= a f(\infty) + b\\ f(\infty)-a f(\infty) &= b\\ f(\infty)\cdot(1-a) &= b\\ f(\infty) &= \frac b{1-a}\\ \end{align} $$ And under this interpretation you get $b/(1-a)$ regardless of the values of $a$ and $b$.
When we see this phenomenom of several answers depening on the interpretation, there is always a clear answer on what the problem is and how to solve it:
The problem is that you have not defined what you mean properly, and the solution is to define what you mean properly.