Does equal Bell series imply Dirichlet convolution? Apostol states in Ch.$2$ , section $2.17$ of his "Intro to Analytic NT" book;

For any two arithmetical Functions $f$ and $g$ let $h = f*g$. Then for every prime $p$ we have $h_p(x)=f_p(x) \cdot g_p(x)$

Let me explain some of the notation in the more common/known (?) setting, here, arithmetical functions just means $f: \mathbb{N} \to \mathbb{R}$ , here, the symbol $*$ is referring to the Dirichlet convolution, and $f_p(x)
$ is referring to the Bell series of $f$ modulo $p$. I know how to prove this, it's quite routine in fact, I want to know if the converse of this holds as well, ie.
If $h_p(x)=f_p(x) \cdot g_p(x)$ for every prime $p$, does this imply $h=f*g$ ?
It is easy to show this is equivalent to asking: If $h(p^n)= f*g \ (p^n)$ for all prime $p$ and $n \in \mathbb{N_{0}}$, does this imply $h=f*g$ ?
I also know that if $f$ and $g$ are multiplicative functions, $f=g \iff f_p(x)=g_p(x)$ for all primes $p$.
So does this mean the answer to my question (in bold) is that the converse holds if (and only if?) $h$ is multiplicative?
I have learnt all this in the past 12 hours only, so I'm not sure if I understand it well at all, but after reading the theorem in Apostol's book, the converse seemed like a natural question to ask, so I did ask myself that question, and I tried to come up with an answer, but I don't think it is satisfactory, mainly because it seems like the thing that the book should have mentioned, as it is very strong(?) and cool and also because I don't think I understand this stuff well right now.
Any answer will be appreciated, Thanks!
 A: Indeed the identity $h_p=f_pg_p$ is equivalent to
$$h(p^n)=(f\ast g)(p^n),$$
for all $n\geq0$. And of course if $f$ and $g$ are multiplicative, this implies $h=f\ast g$.
On the other hand, consider the completely multiplicative functions $g$ and $h$ defined by $g(1)=h(1)=1$ and
$$g(p):=0\qquad\text{ and }\qquad h(p):=-1,$$
for every prime number $p$. The function
$$f(k):=\begin{cases}
(-1)^n&\text{ if $k=p^n$ for some prime $p$ }\\
2&\text{ otherwise}\end{cases}$$
is clearly not multiplicative, as $f(6)=2$ and $f(2)=f(3)=-1$. Then we have
$$(f\ast g)(p^n)=(-1)^n=h(p^n),$$
and so $h_p=f_pg_p$ for every prime number $p$. But for example
$$h(6)=1\qquad\text{ and }\qquad (f\ast g)(6)=2,$$
which shows that $h\neq f\ast g$.
Of course this example can easily be adjusted to have $f$ take any values you like outside the prime powers. The simple fact is that the condition that $h_p=f_pg_p$ sets conditions on $f$, $g$ and $h$ only at the prime powers. Even if $g$ and $h$ are both completely multiplicative, if $f$ is not multiplicative then this tells you absolutely nothing about $f$ outside the prime powers.
A: Ok, I figured this one out myself, infact the converse does not generally hold, but does hold is $f*g$ is multiplicative. As, if $f*g$ is multiplicative and $h$ and $f*g$ agree for all prime powers, they agree for all natural numbers (due to, say, the fundamental theorem of arithmetic).
But if $f*g$ is not multiplicative, then it does not have to be true, I found the following counter example to convince myself.
Consider $h=\frac{\log n}{\Omega(n)}$, $f=\delta$ and $g=\Lambda (n)$,
Where $\Omega(n)$=Prime omega function (big omega) , $\delta$ is the identity for Dirichlet convolution, $\delta(n)= \lfloor\frac{1}{n}\rfloor$ and $\Lambda(n)$ is the Mangoldt function.
Their Bell series are equal for all primes, both are equal to $\frac{\log p}{1-x}$ and indeed they are both take the same value for all powers of primes, but for something like say $n=p^a \cdot q^b$, $h(n) = \frac{\log (p^a \cdot q^b)}{a+b}$ while $f*g(n)=\Lambda(n)=0$
