Note that $20 = 2^2 \cdot 5$.
By Fundamental theorem of Finitely Generated Abelian Group, there are two distinct abelian groups of order $20$: $\mathbb{Z}_{20}$ and $\mathbb{Z}_{10} \times \mathbb{Z}_2$.
Now let $G$ be a nonabelian group of order $20$. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow $5$-subgroup $H \cong \mathbb{Z}_5$. Now let $K$ be any Sylow $2$-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order $20$ is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = \mathbb{Z}_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong \mathbb{Z}_4$; where $\alpha(y) = y^2$.
Let $K = \mathbb{Z}_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$.
If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$.
If $\varphi_2(x) = \alpha$, then $\mathbb{Z}_5 \rtimes_{\varphi_2} \mathbb{Z}_4 $is indeed a nonabelian group of order $20$.
If $\varphi_3(x) = \alpha^2$, then $\mathbb{Z}_5 \rtimes_{\varphi_3} \mathbb{Z}_4$ is indeed a nonabelian group of order $20$. Moreover, since $\mathsf{ker}\ \varphi_3 \cong \mathbb{Z}_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$,$ H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$.
If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $\mathbb{Z}_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$.
Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup.
Suppose now that $K = \mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : \mathbb{Z}_2^2 \rightarrow \mathbb{Z}_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide $2$. We thus have $\psi(a)$, $\psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices.
If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$.
If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $\mathbb{Z}_5 \rtimes_{\psi_2} \mathbb{Z}_2^2$ is indeed a nonabelian group of order $20$.
If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$.
If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\psi_4 = \psi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $\mathbb{Z}_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$.
Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup.
In summary, the distinct groups of order $20$ are as follows. We let $\mathbb{Z}_5 = \langle y \rangle$, $\mathbb{Z}_4 = \langle x \rangle$, and $\mathbb{Z}_2^2 = \langle a \rangle \times \langle b \rangle$.
$Z_{20}$,
$Z_{10} \times Z_2$,
$Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$.
$Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$
$Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$.
(Source: Crazyproject)
Finite abelian groups are determined by the number of elements of any given order. But if you don't know the classification of abelian groups you can always be more concrete in any given case: you can map $(a,b,c) \in (\mathbb{Z}/2 \mathbb{Z})^3$ to $(-1)^a \times 5^b \times 7^c \in (\mathbb{Z}^{\times}_{24})$ and check that this is an isomorphism directly.
As noted in the comments, two groups can have the same number of elements of the same order and not be isomorphic. A good example to keep in mind is the abelian group $G = (\mathbb{Z}/3 \mathbb{Z})^3$ and the non-abelian $3$-Sylow subgroup of $\mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$:
$$\left( \begin{matrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{matrix} \right) \subset \mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$$
Both groups have $27$ elements and every non-trivial element has order $3$. If you don't like the description of that latter group you could think about the permutation group:
$$\langle (1, 4, 7)(2, 5, 8)(3, 6, 9),(1, 3, 2)(4, 5, 6),(4, 6, 5)(7, 8, 9) \rangle \subset S_9.$$
Best Answer
As I have said in my comments, the other answers show that there is a surjective homomorphism $G \to S_4$. To prove that this is an isomorphism, it is enough to prove that $|G| \le 24$, which can be done as follows.
Let $H = \langle a,b \rangle$ be the subgroup of $G$ generated by $a$ and $b$. I am assuming you know that $\langle a,b \mid a^2=b^2=1, aba=bab \rangle$ defines the dihedral group of order $6$, so $|H| \le 6$, and it is enough to prove that $|G:H| \le 4$.
To do that, we shall prove that $G = H \cup Hc \cup Hcb \cup Hcba$, and to do that it is sufficent to prove that if we multiply any of these cosets by any of the generators $a,b,c$ of $G$, then we will get another one of these cosets. So let's do that.
Multiply coset $Hc$ by $a,b,c$: $Hca=Hac=Hc$, $Hcb=Hcb$, $Hcc=H$.
Multiply coset $Hcb$ by $a,b,c$: $Hcba=Hcba$, $Hcbb=Hc$, $Hcbc=Hbcb=Hcb$.
Multiply coset $Hcba$ by $a,b,c$: $Hcbaa=Hcb$, $Hcbab=Hcaba=Hacba=Hcba$, $Hcbac=Hcbca=Hbcba=Hcba$.
This proof generalizes to similar presentations of $S_n$ for all $n$.