It is clear that, being $f$ multiplicative, $f(p^k)f(q^j)=f(p^kq^j)$ whenever $p$ and $q$ are distinct primes. The key observation is that $\frac{f(k)}{k^s}$ is a quotient of two multiplicative functions.
More precisely, for every $k$ on the left hand side consider its factorization: $k=p_1^{a_1}\cdot...\cdot p_n^{a_n}$. Then:
$\frac{f(k)}{k^s}= \frac{f(p_1^{a_1})\cdot...\cdot f(p_n^{a_n})}{p_1^{a_1s}\cdot...\cdot p_n^{a_ns}}= \frac{f(p_1^{a_1})}{p^{a_1s}}\cdot...\cdot\frac{f(p_n^{a_n})}{p_n^{a_ns}}.$
Now you only need to put the pieces together. Since every $k$ on the left hand side has a unique prime factorization on the right hand side the thesis follows.
Let us explicit this argument. The Fundamental Theorem of Arithmetic tells us that there is a $1-1$ correspondence between natural numbers $k\in\mathbb{N}$ and prime factorizations $p_1^{a_1} \cdot ... \cdot p_n^{a_n}$, where the $p_i$ are distinct primes and the $a_i$ are again natural numbers. So we can rewrite the left hand side of your expression as:
$\sum_{k=1}^\infty \frac{f(k)}{k^s} = \sum_{\text{prime factorizations }p_1^{a_1}\cdot...\cdot p_n^{a_n}} \frac{f(p_1^{a_1}\cdot...\cdot p_n^{a_n})}{(p_1^{a_1}\cdot...\cdot p_n^{a_n})^s}.$
How to obtain the sum over prime factorization? We take the product over all the possible factors of the form $p^k$ where $p$ is a prime and $k$ is a natural number: $\{\text{prime factorizations } p_1^{a_1}\cdot...\cdot p_n^{a_n}\}\underset{\text{set}}{=} \{\left(\sum_{k=0}^\infty 2^k\right)\cdot\left(\sum_{k=0}^\infty 3^k\right)\cdot\left(\sum_{k=0}^\infty 5^k\right)\cdot...\}.$
Observe that we must include the factor $p^0=1$ because otherwise we would only have infinite products.
Edit: I believe this kind of argument was first used by Euler to give an analytic proof of the infiniteness of primes. Indeed we know that for every prime $p$ the geometric series: $\sum_{k=1}^\infty \frac{1}{p^k}$ is convergent. While the armonic series $\sum_{n=1}^\infty \frac{1}{n}$ is divergent. But the argument above implies:
$\prod_{p \text{ prime}} \sum_{k=0}^\infty \frac{1}{p^k}= \sum_{n=1}^\infty \frac{1}{n} = \infty.$
So the number of primes must be infinite.