Notice that by the fundamental theorem of finite abelian groups, we have that if $\Gamma$ is abelian and finite that $$\Gamma \cong \bigoplus_{i=1}^n \mathbb{Z}_{p_i^{e_i}}^{k_i} $$ where each of the $p_i$ are primes. Now, if any of the $e_i$ were bigger than one, so $\mathbb{Z}_{p^\alpha}$ is a subgroup of $\Gamma$ with $\alpha$ greater than 1, it follows that the generator of the cyclic group would have order $p^\alpha$ and therefore not have prime order. It follows that for any pleasant group $\Gamma$ we have $$\Gamma \cong \bigoplus_{i=1}^n \mathbb{Z}_{p_i}^{k_i}$$ Now, suppose that this decomposition is composed of at least two different primes $p_1$ and $p_2$, not equal to each other. Then, by properties of the direct sum of groups $\mathbb{Z}_{p_1} \oplus \mathbb{Z}_{p_2}$ is a subgroup of $\Gamma$, and the element that is the product of the generators of these two groups $\mathbb{Z}_{p_1}$ and $\mathbb{Z}_{p_2}$ would have order $p_1p_2$, and therefore cannot be prime. Thus we have for a pleasant group $\Gamma$ $$\Gamma \cong \mathbb{Z}_{p}^k $$ for $p$ prime and $k$ an integer. Furthermore, it can easily be seen each such group is abelian and that the order of each non-identity element in such a group is $p$. Thus pleasant groups are in 1-1 correspondence with prime powers. Therefore, the number of pleasant groups of order at most 25 is simply equal to the number of prime powers less than 25. There are exactly 15 such numbers $$ 1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25$$ which correspond to precisely the groups you have found. And thus it seems that there are 15 pleasant groups of order at most 25. I could be mistaken somewhere in this reasoning but it seems to me that the answers given do not contain the correct choice of 15.
(Note: I just finished teaching a lecture on semi-direct products and their constructions, so my mind went there, especially as I know the structure of the automorphism groups at issue well; but as Steve D points out in comments, you don't even need the automorphism groups and there is a simpler solution that only uses the Sylow theorems, and which I have added below under the horizontal line.)
The only additional thing you need is to know the automorphism groups of the nonabelian groups of order $27$, and of the abelian groups of order $27$.
Let $G$ be a group of order $27p$ with $p$ a prime, $p\gt 3$ and $3\nmid p-1$. The number of $3$-Sylow subgroups of $G$ is congruent to $1$ modulo $3$ and divides $27p$, hence is either $1$ or $p$. But if $3\nmid p-1$, then it cannot be $p$. So the $3$-Sylow subgroup of $G$ is normal. Call it $N$.
Let $C_p$ be a subgroup of order $p$ of $G$ (necessarily cyclic). Then $G$ is a semidirect product of $N$ by $C_p$, since $G=NC_p$ and $N\triangleleft G$. We just need to see what kind of actions $C_p$ can have on $N$.
We have $5$ cases:
$N$ is cyclic of order $27$. The automorphism group of $N$ is cyclic, of order $\phi(27) = 18 = 2\times 3^2$, so the action of $C$ on $N$ is trivial, and we have a direct product, which is cyclic of order $27p$.
$N$ is isomorphic to $C_9\times C_3$, where $C_n$ is the cyclic group of order $n$. As shown here, the automorphism group here has order $\phi(9)\phi(3)3^2 = 4\times 3^4$, and again that means that any map $C_9\to\mathrm{Aut}(C_9\times C_3)$ is trivial, so we have a direct product, which is isomorphic to $C_3\times C_{9p}$.
$N$ is elementary abelian. Then the automorphism group of $N$ is isomorphic to $\mathsf{GL}(3,3)$, which has order $(3^3-1)(3^3-3)(3^3-9) = (26)(24)(18)$. The prime factors are $2$, $3$, and $13$, so you could have a nontrivial semidirect product $N\rtimes C_{p}$ if $p=13$... but in that case $3\mid 13-1$, which we have excluded. For any other prime $p\gt 3$ the action is trivial, so again we get an abelian group, this time isomorphic to $C_3\times C_3\times C_{3p}$.
$N$ is the Heisenberg group of order $27$. As noted here, the automorphism group is isomorphic to
$$\mathsf{AGL}(2,3) = \left\{ \begin{pmatrix}a & b& e\\ c& d& f\\ 0 & 0 & 1\end{pmatrix} : a,b,c,d,e,f \in \mathbb{Z}/3\mathbb{Z},\; ad-bc \neq 0 \right\}.$$
The order of the group is $9$ times the number of choices for $a,b,c,d$. This is equal to the number of ordered pairs of linearly independent vectors in $\mathbb{F}_3^2$, which is $(9-1)(9-3)=2^4\times 3$; so any homomorphism from $C_p$ to $\mathsf{AGL}(2,3)$ is trivial, hence the action of $C_p$ on $N$ is trivial, and $G\cong N\times C_p$.
$N$ is the nonabelian group of order $27$ that has an element of order $9$. Again in Jack Schmidt's post quoted above he notes the automorphism group can be identified with
$$\left\{ \begin{pmatrix}a & b& 0\\ 0& 1& 0\\ c & d & 1\end{pmatrix} : a,b,c,d \in \mathbb{Z}/3\mathbb{Z},\; a ≠ 0 \right\}.$$
This group has order $2\times 3^3$, so any morphism from $C_p$ to the automorphism group of $N$ is trivial and we get $G\cong N\times C_p$.
So the only groups of order $27p$ with $p$ primes, $p\gt 3$, $3\nmid p-1$ are abelian, or of the form $N\times C_p$ where $C_p$ is cyclic of order $p$ and $N$ is nonabelian of order $27$. If you drop the restriction on $3\nmid p-1$ but require the $3$-Sylow to be normal, then you get another nonabelian group by letting $C_{13}$ act on the elementary abelian group of order $27$.
A simpler solution uses only the Sylow theorems, as pointed out by Steve D below. We already know there is a single $3$-Sylow subgroup, so to prove that $G\cong N\times C_p$ it is enough to show there is a single $p$-Sylow subgroup. The number of $p$-Sylow subgroups must divide $27p$ and be congruent to $1$ modulo $p$, so it must divide $27$. The divisors of $27$ greater than $1$ are $3$, $9$, and $27$. The only prime such that $3\equiv 1\pmod{p}$ or $9\equiv 1\pmod{p}$ is $p=2$, which is excluded for being smaller than $3$. The primes for which $27\equiv 1\pmod{p}$ are $2$ and $13$, and $2$ is excluded for being smaller than $3$, and $13$ because $3\mid 13-1$. Thus, there must be a unique Sylow $p$-subgroup as well, so $G\cong N\times C_p$. The abelian $N$ give abelian groups, the nonabelian groups have the desired form.
Best Answer
Here is an expansion of the comment by Derek Holt.
Let $G$ be a group of order $4p$, where $p$ is a prime $>3$ and $\equiv 3\pmod 4$. By Sylow's theorems, we have Sylow subgroups $S_p$ (of order $p$) and $S_2$ (of order $4$) of $G$. The number of such $S_p$ is a divisor of $4p$ and $\equiv 1\pmod p$, hence is $1$, namely $S_p$ is a unique and normal subgroup of order $p$. $S_2$ acts on $S_p$ by $$\varphi\colon S_2\rightarrow \mathrm{Aut}(S_p),\ g\mapsto (x\mapsto gxg^{-1}).$$ Clearly $S_p\cap S_2=\{1\}$ and $S_p$ and $S_2$ generate $G$, so $G$ is a semi-direct product $G\simeq S_p\rtimes S_2$. Note that $S_p=\langle x_1\rangle$ is a cyclic group with $\mathrm{Aut}(S_p)\simeq C_{p-1}$ a cyclic group of order $p-1$ whose generator $\sigma$ maps $x_1$ to $x_1^k$, where $k$ is a primitive root modulo $p$.
Next, we investigate the structure of this semi-direct product. If the map $\varphi$ is trivial, $G$ is abelian (since $S_2$ is abelian), so we omit them. If $S_2=\langle g_1\rangle$ is cyclic, the only nontrivial homomorphism $\varphi$ is $g_1\mapsto \sigma^{(p-1)/2}$ because $p\equiv 3\pmod 4$, and this gives the dicyclic group. If $S_2=\langle g_1,g_2\rangle$ is the $2$-elementary group, then the only nontrivial homomorphism $\varphi$ is such that $g_1\mapsto \mathrm{id}, g_2\mapsto \sigma^{(p-1)/2}$ up to change of generators. This gives the group $D_{4p}\simeq C_2\times D_{2p}$.