This is not an answer to the question, but an expansion of my comment above. I’m going to briefly recap Compton’s method for showing 0–1 laws and limit laws for classes of algebraic structures (in particular, abelian groups). For various reasons, this method is not applicable to noncommutative groups. I’m relying on the excellent presentation in [1], where one can find the details.

We consider classes $\mathcal A$ of finite structures closed under finite direct products with the property that every $A\in\mathcal A$ can be (up to isomorphism) uniquely decomposed as a direct product of indecomposable structures from $\mathcal A$. Let $a(n)$ be the number of nonisomorphic $A\in\mathcal A$ of size $n$, and $A(x)=\sum_{n\le x}a(n)$ (these are the local and global counting functions of $\mathcal A$, respectively). Likewise, if $\mathcal B\subseteq\mathcal A$, let $b(n)$ and $B(x)$ be its local and global counting functions. The goal is to find sufficient conditions to guarantee that whenever $\mathcal B=\{A\in\mathcal A:A\models\phi\}$ for a first-order sentence $\phi$, then the global asymptotic density of $\mathcal B$,
$$\Delta(\mathcal B)=\lim_{n\to\infty}\frac{B(n)}{A(n)},$$
exists (this is called an FO limit law for $\mathcal A$), and ideally, that it is always $0$ or $1$ (an FO 0–1 law).

Under the assumptions above, the isomorphism classes of $\mathcal A$ form a *multiplicative number system*: a free monoid $(\mathsf A,1,\cdot)$ endowed with a norm function $\|\cdot\|\colon\mathsf A\to(\mathbb N^+,1,\cdot)$ which is a monoid homomorphism such that $\|x\|=1$ only if $x=1$. Here, the monoid multiplication is induced by direct product, and the norm is cardinality. The free generators of $\mathsf A$, or *primes*, correspond to the indecomposable algebras $A\in\mathcal A$. Let $\mathsf P$ be the set of primes of $\mathsf A$, 
and $p(n)$ its local counting function. The *Dirichlet generating function* of $\mathsf A$ is
$$\tag{$*$}\mathbf{A}(x)=\sum_{n\ge1}a(n)n^{-x}=\prod_{n\ge2}(1-n^{-x})^{-p(n)}.$$
Finally, a *partition set* is a subset $\mathsf B\subseteq\mathsf A$ that can be written as
$$\mathsf B=\mathsf P_1^{\gamma_1}\cdots\mathsf P_k^{\gamma_k},$$
where $\mathsf P_1\cup\dots\cup\mathsf P_k=\mathsf P$ is a disjoint partition of $\mathsf P$, and each $\gamma_i$ stands for $m_i$, ${\ge}m_i$, or ${\le}m_i$ with $m_i\in\mathbb N$. Here, $\mathsf B\cdot\mathsf C=\{bc:b\in\mathsf B,c\in\mathbf C\}$, $\mathsf B^m=\underbrace{\mathsf B\cdots\mathsf B}_{m}$, $\mathsf B^{\ge m}=\bigcup_{n\ge m}\mathsf B^n$, and likewise for ${\le}m$.

Now, the strategy for proving FO limit and 0–1 laws goes as follows:

1. If the Dirichlet series $\mathbf A(x)$ has a finite abscissa of convergence $\alpha<\infty$, then every partition set $\mathsf B$ has a *Dirichlet density*
$$\partial(\mathsf B)=\lim_{x\to\alpha+}\frac{\mathbf B(x)}{\mathbf A(x)},$$
where $\mathbf B(x)$ is the generating function of $\mathsf B$.

2. If $A(n)$ satisfies some regularity conditions (which hold e.g. if $A(n)\sim cn^\alpha$), then one can prove a Tauberian theorem showing that every partition set has a global asymptotic density, which agrees with its Dirichlet density. Under some conditions, the density also turns out to be always 0 or 1.

3. By the Feferman–Vaught theorem, the truth of any FO sentence in a direct product $\prod_{i\in I}A_i$ is equivalent to $\mathcal P(I)\models\Phi([[\phi_1]],\dots,[[\phi_k]])$, where $\Phi$ is a formula in the language of Boolean algebras, $\phi_1,\dots,\phi_k$ are sentences in the language of $\phi$, and $[[\phi]]=\{i\in I:A_i\models\phi\}$. Using elimination of quantifiers for atomic Boolean algebras, one can further make $\Phi$ a propositional combination of formulas asserting $|[[\phi_j]]|\ge m_j$ for integer constants $m_j$. This can be used to show that in $\mathcal A$, every FO sentence defines a partition set. Thus, if $\alpha<\infty$ and $A(n)$ satisfies the conditions from 2, $\mathcal A$ has an FO limit law, or even a 0–1 law.

This machinery works well for abelian groups, and shows that finite abelian groups have an FO limit law, and for a fixed prime $p$, abelian $p$-groups have a 0–1 law. One can also readily see that abelian groups do not have a 0–1 law. Note that abelian groups have
$$p(n)=\begin{cases}1&\text{if $n$ is a prime power,}\\0&\text{otherwise,}\end{cases}$$
hence by $(*)$, their Dirichlet generating function is
$$\mathbf A(x)=\prod_{n\ge1}\zeta(nx).$$
The class $\mathcal B$ of abelian groups of odd order has a similar generating function but with the terms for powers of $2$ removed. Since the abscissa of convergence is $\alpha=1$ here, it follows easily that the Dirichlet density of $\mathcal B$ relative to $\mathcal A$ is
$$\prod_{n\ge1}(1-2^{-n})\approx0{.}71,$$
and by the general results, this is also its global asymptotic density. $\mathcal B$ is first-order definable, as it consists of abelian groups with no element of order $2$.

Unfortunately, this strategy does not work for noncommutative groups. For one thing, the analytic methods using Dirichlet generating series rely on the condition that the series converges at least somewhere, i.e., that the abscissa of convergence is finite. This is equivalent to the condition that $a(n)$ is polynomially bounded. However, by results quoted above in the comments, the number of groups of order $2^n$ is $2^{\tfrac2{27}n^3+O(n^{8/3})}$, which grows much too fast. What is even worse is that if we assume that almost all groups are $2$-groups, and that the number of $2$-groups grows smoothly enough (the estimate above is not precise enough for this), then in fact almost all  groups are directly indecomposable, which turns the whole approach on its head: if we deal only with indecomposable structures, then the fact that every formula defines a partition set carries zero information, and there is no way that all sets of indecomposable structures could have asymptotic density.

**Reference:**

[1] Stanley N. Burris, *Number theoretic density and logical limit laws*, Mathematical Surveys and Monographs vol. 86, AMS, 2001, xx+289 pp.