Let $G$ be a group in question. First note that $G/Z(G)\cong C_{p^2}$ isn't possible. Thus $G$ has the presentation $$\langle Z,x,y\mid Z \text{ central}, x^p=a,y^p=b,[x,y]=c\rangle$$ where $Z$ is the center, $a,b, c \in Z$ and $c\neq 1,\; c^p=1$.
Added: Suppose $a=a_0^p,a_0 \in Z$. By replacing $x$ by $xa_0^{-1}$ we have the relation $x^p = 1$. Write $Z=\langle z_1,...,z_n\mid r(Z)\rangle$. Then we obtain the presentations $$\tag{I}G(c)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=y^p=1,[x,y]=c\rangle$$
$$G(c,i)= \langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i, y^p=1,[x,y]=c\rangle\tag{II}$$
$$G(c,i,j)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i,y^p=z_j,[x,y]=c\rangle\tag{III}$$
Clearly, $G(c,-) \cong G(c',-)$ if $\langle c\rangle = \langle c'\rangle$ and $G(c,i) \cong G(c,j)$ if $\exp(z_i)=\exp(z_j)$ and similar for $G(c,i,j)$.
Added 2: 1) In case (III) we can assume $i\neq j$ (otherwise, replacing $x$ by $xy^{-1}$ gives case (II)).
- Let $\exp(z_i)=k_i$. Denote by $(k_1,...,k_n)$ the isomorphism type of $C_{p^{k_1}} \times \cdots C_{p^{k_n}}$. Then the groups above have maximal abelian subgroups of the following types: $$\begin{array}{lcl} (I) & : & (k_1,...,k_n,1) \newline (II) & : & (k_1,...,k_n,1), (k_1,..,k_i+1,..,k_n) \newline (III) & : & (k_1,..,k_i+1,..,k_n), (k_1,..,k_j+1,..,k_n) \end{array}$$ Hence (I), (II), (III) belong to different ismorphism types.
It remains to check for which parameters $c,i,j$ the groups within (I) resp. (II) resp. (III) are isomorphic.