Introduction
Ok, so you want a more detailed answer to this question.
Let me explain :
On your attempt
First, a comment on your attempt.
Note that you don't get an indeterminate form : a constant times zero, is zero. An indeterminate form is when you are multiplying something whose limit is infinity versus something going to zero : only then can the "multiplication rule for limits" fail. It doesn't fail if both limits are real numbers, because in that case their product makes sense.
So your reasoning is spot on : but it's only a specific case.
Understanding the problem
Let us start with what a multiplicative function is : it's an arithmetic function $f$ such that $f(mn) = f(m)f(n)$ for all $m,n \in \mathbb N$ coprime.
The most important thing about an arithmetic function, is to note that it's determined by its values at prime powers. In other words, we know that :
$$
f(\prod_{i=1}^n p_i^{\alpha_i}) = \prod_{i=1}^n f(p_i^{\alpha_i})
$$
where the $\alpha_i>0$ are positive integers and the $p_i$ are distinct primes.
So it's not completely unexpected, that the given statement is true. If the building blocks have a pattern, they build to form a more general pattern.
However, what isn't the right approach, is the kind of approach being suggested by the attempt. The point is, that there are infinitely many primes, so studying particular sequences of primes is not going to resolve the situation, because you are going to leave out the "rest" of the primes in that process.
So we need to go by definition.
That is , we should show that :
For every $\epsilon>0$ there is an $N$ such that $n>N$ implies $f(n)<\epsilon$.
For ease of speech, I call $f(n)$ as the $f$-value of $n$ in some parts of this answer.
The key ideas
Let us first decode the given assumption. While it was worded a little too ambiguously for my liking, Apostol clarifies it to be :
For all $\epsilon>0$ there is an $N>0$ such that whenever $p$ is prime, $m$ is a positive integer such that $p^m > N$, we have $f(p^m)<\epsilon$.
Now, this is interesting. This is basically telling you : if you give me $\epsilon$, I'll give you an $N$ such that all prime powers after $N$ are handled by $\epsilon$. You need to tell me how to improve this $N$ so that all numbers, and not just prime powers, are handled.
But the point is simple : how do we handle numbers that aren't prime powers? Well, let's take a number that's not a prime power. It's definitely a product of prime powers, though!
And now, let's talk about those prime powers in that product. If a prime power is larger than $N$, then it's controlled for us. If it's not larger than $N$, then thankfully it's part of only a finite bunch of prime powers that sit below $N$! Finiteness allows us to take the worst-case and adjust $N$ so that it gets bounded as nicely as needed. And that, my friend, is the game in a nutshell.
The worst case
Take $\epsilon=1$. Recalling the definition of the convergence Apostol is referring to , I quote :
For all $\epsilon>0$ there is an $N>0$ such that whenever $p$ is prime, $m$ is a positive integer such that $p^m > N$, we have $f(p^m)<\epsilon$.
Now, let $\epsilon=1$. There is then a $D>0$ such that whenever $p$ is prime and $m$ is a positive integer such that $p^m>D$, we have $|f(p^m)| < 1$. We just used the definition above to confirm this (note that this definition is given in the same question in Apostol, at least in the version I'm consulting).
The point is, that if a prime power has a high $f$-value, then it must be smaller than $D$. In other words, our worst-case scenario must come from prime powers below $D$. Let $S' = \{n = p^m :n \leq D\}$ be the set of all prime powers below $D$. Define :
$$
T = \max_{S'' \subset S'} \prod_{s \in S''} |f(s)|.
$$
then $T$ is a finite constant that captures the worst possible behaviour away from $D$, in the sense that any number which has prime factors solely consisting of prime factors smaller than $D$, can have $f$-value at most $T$. Thus, we've controlled out worst-case factor.
Now, any prime power bigger than $D$ will only bring down the absolute value if multiplied with another number, because it's $f$-value has modulus smaller than $1$.
Key principle and explanation for key claim
Now, let's look at a term like $p_1^{\alpha_1}...p_m^{\alpha_m}$. Let's imagine that one of the $|f(p_l^{\alpha_l})|$ was smaller than $\epsilon$, and the rest are all smaller than, say $1$. Then,$|f|$ applied to the entire term is smaller than $\epsilon$ by the multiplicative property. We realize from this simple bit of thought, the key principle that one prime power having controlled $f$-value controls the $f$-value of the entire number, since the rest of the prime powers are already under control.
Let's keep ourselves flexible and let $\epsilon>0$. We will also consider a $1>\delta>0$ (note : the $1>$ part is also important) that we will allow to depend upon $\epsilon$, and for this $\delta$ let's get an $N_{\delta}$ such that $p^m > N_{\delta}$ implies $|f(p^m)|<\delta$.
So what does this mean? We have controlled to high degree, the value of $f$ for prime powers above $N_{\delta}$. We ask ourselves in line with the key principle : is it true, that every large enough number is divisible by some prime power above $N_{\delta}$? The answer to that is the key to the entire process, and is an absolute YES.
The idea is to take the product of all prime powers below $N_{\delta}$. If a number is above that product, then it will be divisible by at least one prime power which is bigger than $N_{\delta}$ , as we show below.
Key claim
Claim : Let $P = \{n=p^m, n \leq N_{\delta}\}$ and let $K = \prod_{s \in P} s$. Then, for all $L>K$ there exists a prime power not in $P$ which divides $L$.
Proof : Suppose not. Then, every prime power which divides $L$ belongs in $P$. Now, $L$ is made of a product of prime powers, but each such prime power must be in $P$, so there is some subset $S' \subset P$ such that $L = \prod_{s \in S'} s$. But that clearly means that $L$ divides $K$! So $L \leq K$, a contradiction.$\blacksquare$
What we've basically found, is a constant $K$ such that every number above $K$ has a prime power factor whose $f$-value is well controlled.
Grouping the prime powers
The key principle, combined with the claim, tells us that every large enough number can be controlled, because it contains one controlling factor, and one controlling factor is enough.
Let us now take this to the next level. Remember $D$? We had defined $D$ such that whenever $p^m>D$ we have $|f(p^m)|<1$.
If we have a number $M$, we can consider this decomposition into three parts :
Those coming from prime powers that lie below $D$.
Those coming from prime powers that lie between $D$ and $K$ ($D>K$ could happen, but for the same of rigour we note that the claim works if $K$ is replaced by a larger number as well, so we can take $K$ to be large enough in the claim to ensure that $D<K$ and hence this range makes sense).
Those coming from prime powers that lie above $K$.
Now, the first group contributes at most $T$ in absolute value. The second group, if it exists at all, consists of prime powers $p^m$ such that $|f(p^m)|<1$, so if anything they will reduce the modulus.
The third group contributes at most $\delta$ : this is because if you break it into prime powers, then each prime power has $f$-value at most $\delta$, so if there are $m$ prime powers (and by the claim, $m \geq 1$) then the $f$-value of this group is at most $\delta^m \leq \delta$ because $\delta<1$.
Finally, we conclude that the total contribution is at most $\delta T$. Of course, for this to be less than $\epsilon$, we must have $\delta < \frac{\epsilon}{T}$, and we can now finish the formal proof!
Formal Proof
Let $\epsilon>0$. Let $\delta = \frac{\epsilon}{T}$. Consider $N_{\delta}$ and $K$ defined above corresponding to $\delta$. We claim that for any $L>K$ we have $|f(L)|<\epsilon$.
To see why, we write $L = A \times B \times C$, where :
$A$ is the product of the prime powers dividing $L$, which are smaller than $D$.
$B$ is the product of the prime powers dividing $L$ which are between $K$ and $D$.
$C$ is the product of the prime powers dividing $L$ which are bigger than $K$.
We know that $f(L)=f(A)f(B)f(C)$. Since $C>1$, it follows that $|f(C)|<\delta$. We know that $|f(B)|<1$ and $|f(A)|\leq T$. Therefore, we get that $|f(L)| \leq \delta T = \epsilon$, completing the proof.
Best Answer
What is unclear to me is how you go from $(fg)^{-1}$ to $f^{-1}g^{-1}$. Given that we are dealing with Dirichlet inverses, this is not obvious to me. In fact, it's not true. For example, if $f=g=u$, then $(fg)^{-1} = u^{-1} = \mu$, but $f^{-1}g^{-1} = \mu\mu\neq \mu$.
So that is where the error in your reasoning comes, I think.
Remember that since ${}^{-1}$ represents the Dirichlet inverse, which is the inverse of convolution, you know that $$(f*g)^{-1} = g^{-1}*f^{-1} = f^{-1}*g^{-1},$$ but you are trying to apply this to the pointwise product, which is incorrect.
As to proving the desired identity, if $h$ is an arithmetic function with $h(1)\neq 0$, then $h^{-1}$ is given by: $$ h^{-1}(1) = \frac{1}{h(1)},\qquad h^{-1}(n) = \frac{-1}{h(1)}\sum_{{d|n},\,{d\lt n}}h\left(\frac{n}{d}\right)h^{-1}(d),\quad n\gt 1.$$
If $h$ is completely multiplicative then you have $f^{-1}(n)=\mu(n)f(n)$ for all $n\geq 1$.
So we verify that the identity holds for $f$ completely multiplicative and $g$ an arithmetic function with $g(1)\neq 0$.
At $n=1$, we have $$(f\cdot g)^{-1}(1) = \frac{1}{f(1)g(1)} = \frac{1}{g(1)}$$ (since $f(1)=1$ must hold); and $$(f\cdot g^{-1})(1) = f(1)g^{-1}(1) = \frac{1}{g(1)}.$$
If $n\gt 1$ and the result hold for all integers smaller than $n$, then we have: $$\begin{align*} (f\cdot g)^{-1}(n) &= \frac{-1}{(f\cdot g)(1)} \sum_{d|n, d\lt n}(f\cdot g)\left(\frac{n}{d}\right)(f\cdot g)^{-1}(d)\\ &= \frac{-1}{g(1)} \sum_{d|n\,d\lt n} f\left(\frac{n}{d}\right)g\left(\frac{n}{d}\right) f(d)g^{-1}(d)\\ &= \frac{-1}{g(1)}\sum_{d|n\,d\lt n}\frac{f(n)}{f(d)}g\left(\frac{n}{d}\right)f(d)g^{-1}(d)\\ &= \frac{-f(n)}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac {n}{d}\right)g^{-1}(d)\\ &= f(n)\left(-\frac{1}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac{n}{d}\right)g^{-1}(d)\right)\\ &= f(n)g^{-1}(n)\\ &= (f\cdot g^{-1})(n). \end{align*}$$ We used the fact that $f$ is completely multiplicative to get that $f\left(\frac{n}{d}\right) = \frac{f(n)}{f(d)}$. Equivalently, you can note that $f\left(\frac{n}{d}\right)f(d) = f(n)$.
This proves the first part of the exercise.
The second part of the exercise asks you to show that if $f$ is multiplicative and $$(f\cdot \mu^{-1})^{-1} = f\cdot \mu$$ holds, then $f$ is completely multiplicative.
Since $f$ is completely multiplicative if and only if $f^{-1} = \mu\cdot f$, you are being asked to show that $f\cdot \mu^{-1} = f$. That is, you need to show that $\mu^{-1}(n) = 1 = u(n)$.
But we know this, since $\sum_{d|n}\mu(d) = I(n)$ (the identity of the convolution), so $u$ and $\mu$ are inverses of each other.