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.