For each prime there are exactly two nonabelian groups of order $p^3$ up to isomorphism

$P$ is extraspecial: $Z(P)=[P,P]=\mathrm{\Phi }(P)=⟨z⟩$ has order $p$, and $V=P\mathrm{/}Z(P)$ is elementary abelian of order $p^2$.

Since $\mathrm{\Phi }(P)=[P,P]P^p$ contains $P^p$, every $g\in P$ has $g^p\in Z(P)$.

Every element order divides $p^3$; an element of order $p^3$ would generate $P$ and make it abelian, so every element has order dividing $p^2$ and $\exp (P)$ is $p$ or $p^2$.

For odd $p$ the groups $H_p$ and $M_p$ are nonabelian of order $p^3$ with exponents $p$ and $p^2$, so they are not isomorphic.

The groups $\mathrm{Dih}(C_4)$ and $Q_8$ are nonabelian of order eight and have six and two solutions of $x^2=1$, so they are not isomorphic.

First case: suppose $g^p=e$ for every $g\in P$. At $p=2$ this makes $P$ abelian, since $gh=(gh{)}^{-1}=h^{-1}{g}^{-1}=hg$; so $p$ is odd here.

Second case: suppose instead that $P$ has an element $a$ of order $p^2$. Then $⟨a⟩$ has order $p^2$ and index $p$, and $a^p$ is a nonidentity element of $Z(P)$, so $Z(P)=⟨a^p⟩$.

In the first case, nonabelianness gives $x,y$ with $[x,y]\mathrm{\ne }e$; then $z:=[x,y]$ generates $Z(P)$ and $⟨x,y⟩$ has order $p^3$, so $P=⟨x,y⟩$.

In the second case, choose $b\mathrm{\notin }⟨a⟩$; then $⟨a,b⟩$ properly contains a subgroup of index $p$ and so equals $P$, and $[b,a]\mathrm{\ne }e$ since otherwise $P$ would be abelian.

In the first case every element of $P$ is uniquely $x^u{y}^v{z}^w$ with $u,v,w$ in $\{0,\dots ,p-1\}$: the images $\bar{x},\bar{y}$ generate $V$, so the products $x^u{y}^v$ meet every coset of $Z(P)=\{z^w\}$, and there are exactly $p^3$ such triples of exponents.

In the second case $[b,a]$ lies in $⟨a^p⟩$ and is not the identity, so $[b,a]=a^{pk}$ with $p\nmid k$; choosing $k^{\mathrm{\prime }}$ with $k{k}^{\mathrm{\prime }}\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and replacing $b$ by $b^{k^{\mathrm{\prime }}}$, which still lies outside $⟨a⟩$ because its image in $V$ is nontrivial, gives $[b,a]=a^p$, that is $ba{b}^{-1}=a^{1+p}$. Moreover $b^p\in Z(P)=⟨a^p⟩$, say $b^p=a^{pm}$.

In the first case the class-two identities give $y^v{x}^{u^{\mathrm{\prime }}}=[y^v,x^{u^{\mathrm{\prime }}}]x^{u^{\mathrm{\prime }}}{y}^v=z^{-u^{\mathrm{\prime }}v}{x}^{u^{\mathrm{\prime }}}{y}^v$, so $(x^u{y}^v{z}^w)(x^{u^{\mathrm{\prime }}}{y}^{v^{\mathrm{\prime }}}{z}^{w^{\mathrm{\prime }}})=x^{u+u^{\mathrm{\prime }}}{y}^{v+v^{\mathrm{\prime }}}{z}^{w+w^{\mathrm{\prime }}-u^{\mathrm{\prime }}v}$, all exponents read modulo $p$ because $x^p=y^p=z^p=e$.

In the second case with $p$ odd, the $p$-th power map is a homomorphism, so $c=b{a}^{-m}$ has $c^p=b^p(a^{-m}{)}^p=e$; also $c\mathrm{\notin }⟨a⟩$, and $ca{c}^{-1}=ba{b}^{-1}=a^{1+p}$.

In the second case with $p=2$, the element $b^2$ lies in $\{e,a^2\}$, and the relation reads $ba{b}^{-1}=a^3=a^{-1}$.

In $H_p$ put $X=(1,0,0)$, $Y=(0,1,0)$ and $W=(0,0,1)$. Then $XY=(1,1,1)$, $YX=(1,1,0)$ and $(YX{)}^{-1}=(-1,-1,1)$, so $[X,Y]=(XY)(YX{)}^{-1}=(0,0,1)=W$, which generates $Z(H_p)$; for odd $p$ every element of $H_p$ has $p$-th power the identity, so steps 3.1 and 4.1 hold verbatim in $H_p$ with $X,Y,W$ in place of $x,y,z$.

In the second case with $p$ odd, $⟨a⟩$ is normal because $c$ conjugates it into itself and $a$ normalises it, $⟨a⟩\cap ⟨c⟩$ is trivial because $c\mathrm{\notin }⟨a⟩$ and $⟨c⟩$ has prime order, and $\mid ⟨a⟩\mid \mid ⟨c⟩\mid =p^3$, so $P=⟨a⟩⟨c⟩$ is an internal semidirect product whose conjugation action sends $a$ to $a^{1+p}$; hence $P\cong M_p$.

In the second case with $p=2$ and $b^2=e$: $⟨a⟩$ is normal of index two, $⟨a⟩\cap ⟨b⟩$ is trivial, and $b$ acts on $⟨a⟩$ by inversion, so $P$ is the internal semidirect product of a cyclic group of order four by a cyclic group of order two acting by inversion, that is $P\cong \mathrm{Dih}(C_4)$.

In the second case with $p=2$ and $b^2=a^2$: the eight elements $a^u{b}^v$ with $0\le u<4$ and $v\in \{0,1\}$ are distinct and exhaust $P$, and the relations $a^4=e$, $b^2=a^2$ and $ba=a^{-1}b$ determine every product of two of them. The quaternion group satisfies the same three relations with $\mathbf{i}$ for $a$ and $\mathbf{j}$ for $b$, since ${\mathbf{i}}^4=1$, ${\mathbf{j}}^2=-1={\mathbf{i}}^2$ and $\mathbf{j}\mathbf{i}=-\mathbf{k}={\mathbf{i}}^{-1}\mathbf{j}$, and $\mathbf{j}\mathrm{\notin }⟨\mathbf{i}⟩$, so its eight elements have the same normal form; matching normal forms is therefore an isomorphism and $P\cong Q_8$.

In the first case, matching normal forms gives a bijection $H_p\to P$ carrying $X^u{Y}^v{W}^w$ to $x^u{y}^v{z}^w$, and both products are computed by the same rule, so it is an isomorphism and $P\cong H_p$.

The two cases are exhaustive, and within the second the two parities are exhaustive; so for odd $p$ every nonabelian group of order $p^3$ is isomorphic to $H_p$ or to $M_p$, and for $p=2$ to $\mathrm{Dih}(C_4)$ or to $Q_8$. With the two non-isomorphy statements this gives exactly two isomorphism classes at every prime.