For your first question: let $R$ be any commutative ring, and let $D(R)$ be the ring of formal Dirichlet series over $R$, i.e., the set of all functions $f: \mathbb{Z}^+ \rightarrow R$ under pointwise addition and convolution product.
Then the unit group of $R$ is precisely the set of formal Dirichlet series $f$ such that $f(1)$ is a unit in $R$.
As for your second question, it is indeed equivalent to asking whether $U(D(R))$ is $n$-divisible. Here, if we take $R = \mathbb{Z}$ as you asked, the answer is that for all $n \geq 2$, $U(D(\mathbb{Z}))$ is not $n$-divisible and that even the Dirichlet series $\zeta(s)$ is not an $n$th power in $D(\mathbb{Z})$.
[Now, for some reason, I switch back to the classical notation, i.e., I replace the arithmetical function $f$ by its "Dirichlet generating series" $\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$. It would have been simpler not to do this, but too late.]
Let $f(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s}$ be any formal Dirichlet series, and suppose that $g(s) = \sum_{n=1}^{\infty} \frac{b_n}{n^s}$ be a formal Dirichlet series such that $g^2 = f$. Thus
$a_1 + \frac{a_2}{2^s} + \ldots = (b_1 + \frac{b_2}{2^s} + ... )(b_1 + \frac{b_2}{2^s} + \ldots)$
$= b_1^2 + \frac{2 b_1 b_2}{2^s} + \frac{2 b_1 b_3}{3^s} + \frac{2b_1 b_4 + b_2^2}{4^s} + \ldots$
(This multiplication is formal, i.e., it is true by definition.)
Thus $b_1 = \pm \sqrt{a_1}$. Suppose we take the plus sign, for simplicity. Then for all primes $p$,
$a_p = 2 b_1 b_p$, so
$b_p = \frac{a_p}{2 \sqrt{a_1} }$,
so we need $2 \sqrt{a_1}$ to divide $a_p$, so at least we need $a_p$ to be even for all primes $p$. Further conditions will come from the composite terms.
These same considerations show that if we replaced the coefficient ring $\mathbb{Z}$ by $\mathbb{Q}$ (or any coefficient field of characteristic $0$), then any formal Dirichlet series with $a_1 = 1$ is $n$-divisible for all positive integers $n$. In particular, you can write $\zeta(s)^{\frac{1}{n}}$ as a Dirichlet series with $\mathbb{Q}$-coefficients just by applying the above procedure and successively solving for the coefficients. Whether there is a nice formula for these coefficients is a question for a better combinatorialist than I to answer.
EDIT: Based on your comments below, I now understand that you are looking for a characterization of $U(D(\mathbb{Z}))$ as an astract abelian group. I believe it is isomorphic to
$\{ \pm 1 \} \times \prod_{i=1}^{\infty} \mathbb{Z}$. (Or, more transparently, to
the product of ${ \pm 1}$ with the product of infinitely many copies of $\prod_{i=1}^{\infty} \mathbb{Z}$, one for each prime number. But as abstract groups it amounts to the same thing.)
