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 
$$G(c)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=y^p=1,[x,y]=c\rangle$$

$$G(c,i)= \langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i, y^p=1,[x,y]=c\rangle$$

$$G(c,i,j)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i,y^p=z_j,[x,y]=c\rangle$$

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 $G(c,i,j)$. 

I don't know at the moment, if $G(c), G(c,i),G(c,i,j)$ have different isomorphism types. But I'll think about it.