zeta functions - Fractional powers of Dirichlet series?

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.)