A possible characterization of the category of finite $p$-groups
$\DeclareMathOperator\GL{GL}\newcommand\card[1]{\lvert#1\rvert}$Here is a proof of "yes," using Tim Campion's proposition below.
Let $p$ be the smallest prime in $P$. For any prime $q\ne p$, let $o_q(p)$ be the multiplicative order of $p$ modulo $q$, or equivalently the least $n$ such that $q$ divides $\card{\GL_n(p)}$. Assuming $P\ne\{p\}$, let $q\in P-\{p\}$ and $G=\GL(V)$ where $V$ is a vector space of order $p^n$ and $n=o_q(p)$. As $p$ is smallest, $n\ge2$. By assumption, $G$ contains a Hall $P$-subgroup $X$. Then $X$ contains a Sylow $p$-subgroup $U$ of $G$ as well as an element $x$ of order $q$. If $U$ is normal in $X$, then $\card X$ divides $\card U(p-1)^n$, the order of the full upper triangular group, which is not divisible by $q$ as $q>p$. So $U$ is not normal in $X$. The theory of $B$-$N$ pairs then implies that $X$ contains a copy of $\operatorname{SL}_2(p)$. Hence every prime divisor of $p-1$ lies in $P$, which forces $p=2$. Then $U$ is a Borel subgroup of $G$, so $X$ must be a parabolic subgroup of $G$. But because of $x$, $X$ stabilizes no proper subspace of $V$. The only such parabolic subgroup is $X=G$. Hence $P$ contains all prime divisors of $\card G$. In particular, $3\in P$.
Now suppose that $P$ is not the set of all primes and choose a prime $r\not\in P$ to minimize $m=o_r(2)$. Since $3\in P$, $m\ge3$. Let $H=\GL(W)$, where $W$ is a vector space of order $2^m$. Then $H$ contains an element $y$ of order $r$. Let $U$ be a Sylow $2$-subgroup of $H$. Let $W_1$ and $W_{m-1}$ be $U$-invariant subspaces of $W$ of respective dimensions $1$ and $m-1$. Let $H_1$ and $H_{m-1}$ be their respective stabilizers in $H$. Then $H_1$ and $H_{m-1}$ are maximal parabolic subgroups of $H$ containing $U$, and each is an extension of an elementary abelian $2$-group by $\GL_{m-1}(2)$. Hence $H_1$ and $H_{m-1}$ are $P$-groups, by our choice of $r$. However, they are maximal subgroups of $H$ and they are not conjugate in $H$, being distinct parabolic subgroups containing $U$.
By assumption, $\langle H_1^g, H_{m-1}\rangle$ must be a $P$-group for some $g\in H$. Since $H_{m-1}$ is maximal and not equal to $H_1^g$, $H$ must be a $P$-group. But $y\in H$ has order $r\notin P$, contradiction.
We can get most of the way to an answer, without using all the conditions.
Definition: Say that $S \subseteq \mathcal{FG}$ is nice if it is nonempty and closed under subgroups, extensions, and isomorphy, and has the property that for any $G \in \mathcal{FG}$, the $S$-maximal subgroups of $G$ are all conjugate.
Proposition: Suppose that $S \subseteq \mathcal{FG}$ is nice. Then there is a set $\mathcal P$ of primes such that $P \in S$ if and only if the prime divisors of $|P|$ are all in $\mathcal P$.
This implies that any $\mathcal P$-Hall subgroup of $G$ is a maximal $S$-subgroup of $G$. Conversely, if $G$ has a maximal $S$-subgroup $H$, then assuming niceness, since $H$ contains the Sylow $p$-subgroups for each $p \in \mathcal P$, we have that $H$ is a $\mathcal P$-Hall subgroup. That is, if $S$ is nice, then a maximal $S$-subgroup is precisely a Hall $\mathcal P$-subgroup.
Proof: Let $\mathcal P$ be the set of primes $p$ such that there exists $P \in S$ with $p$ dividing $|P|$ (as it must be). By passing to subgroups, we have $C_p \in S$ for $p \in \mathcal P$, and by taking extensions we have that every finite $p$-group is in $S$ for every $p \in \mathcal P$. Let $G$ be a finite group such that the prime divisors of $|G|$ are in $\mathcal P$. Now, there are $S$-maximal subgroups of $G$ containing each Sylow subgroup of $G$, and by hypothesis these are all conjugate and in particular have the same order. So if $|G| = p^n m$ with $p$ not dividing $m$, then $p^n$ divides the order of any $S$-maximal subgroup $P \subseteq G$. Since this is true for all prime divisors $p$ of $|G|$, we have that $|G|$ divides $|P|$, so that $P = G$, and thus $G \in S$.
It's known that not all $\mathcal P$-Hall subgroups of a finite group $G$ are conjugate in general, but I don't happen to know whether the known counterexamples cover all possible sets $\mathcal P$ of primes other than the subsingletons and the set of all primes. If $G$ is solvable, then its $\mathcal P$-Hall subgroups are conjugate -- so if we replace $\mathcal{FG}$ with just the solvable finite groups we get an expanded set of possibilities for $S$.
(Many of the comments relate to an earlier version of this post which attempted to whittle down the set of $\mathcal P$ allowed. Thanks to Richard Lyons below for pointing out the mistake.)
EDIT: I just wanted to add another way to close the argument, alternate to Richard Lyons' above, which relies on the original literature on Hall subgroups.
Hall showed in Theorems Like Sylow's, discussion after Theorem A4, that if $\mathcal P$ contains two primes smaller than $n$, then any Hall $\mathcal P$-subgroup of $\Sigma_n$ cannot be solvable. Then Thompson showed in Hall subgroups of the symmetric groups that the only nonsolvable Hall subgroups of $\Sigma_n$ are the trivial group, $\Sigma_n$ itself, and (if $n$ is prime) $\Sigma_{n-1}$. So if $\mathcal P$ contains $p,q$ and omits $r$, then take $G = \Sigma_n$ for any composite $n > p,q,r$. The only candidate Hall subgroups of $G$ are the trivial group and $G$ itself; the former is not a $\mathcal P$-Hall subgroup because $p | n!$ and the latter is not because $r | n!$.