I have two questions regarding to $p$-groups.
Every $p$-group is a homomorphic image of a $p$-group with cyclic center of order $p$, so a classification (whatever that means) of $p$-groups with cyclic center would (more-or-less) yield a construction for all $p$-groups, and I would not hold my breath waiting for that.
To see why a $p$-group $P$ is a homomorphic image of a $p$-group $G$ with center of order $p$, let $G$ be the regular wreath product of a cyclic group of order $p$ with $P$. Thus $G$ has an elementary abelian subgroup $E$ of order $p^{|P|}$, where $P$ permutes the cyclic factors of $E$ the way it permutes its own elements by right multiplication, and thus $P$ acts faithfully on $E$. Also, $G$ is the semidirect product of $E$ by $P$. It is easy to see that $E \cap {\bf Z}(G)$ has order $p$, so I need to show that every element of ${\bf Z}(G)$ lies in $E$. If $z \in {\bf Z}(G)$, write $z = au$, where $a \in E$ and $u \in P$. Since $z$ centralizes $E$ and $a$ centralizes $E$, it follows that $u$ centralizes $E$ and thus $u = 1$ by the faithfulness of the action. Thus $z = a \in E$, as required.
This was mentioned as a brief comment by Derek Holt, but I think it deserves to be an answer. For a group of order $p^n$ and class $c$, the coclass is $n-c$. The known theory of classification by coclass is much richer than the known theory of classification by class; there is a nice account in the paper On the classification of prime-power groups by coclass by Eick and Leedham-Green (also available here). Part of the story is that one can also define coclass for infinite pro-$p$-groups, which is useful for the classification. The isomorphism types of finite $p$-groups of fixed coclass can be assembled into a tree in a natural way, and the infinite paths from the root of the tree correspond to isomorphism classes of pro-$p$-groups.
To echo Marty's sentiment, a classification of $p$-groups based on nilpotency class and size seems far off. As an example of where this problem can get complicated, see
http://www.degruyter.com/view/j/jgth.2011.14.issue-6/jgt.2010.081/jgt.2010.081.xml.
Basically, Halasi and Palfy construct a collection of $p$-groups (allowing $p$ to vary) of nilpotency class 2 where the number of conjugacy classes is not polynomial in the prime $p$. This is related to the (unsolved) question of whether the number of conjugacy classes of $UT_n(\mathbb{F}_q)$, the group of unipotent upper-triangular matrices over the field with $q$ elements, is polynomial in $q$.
To summarize, $p$-groups, even those of nilpotency class 2, get complicated.
This is not quite what you were asking, since it involves the lower central series instead of the upper one, but Miech classified the $2$-generator $p$-groups with cyclic commutator subgroup and $p$ odd in:
Miech, R.J. On $p$-groups with a cyclic commutator subgroup J. Austral. Math. Soc. 20 no. 2 (1975), 178-198, MR0404441 (53 #8243).
Their nilpotency class can be arbitrarily large, though of course they are all metabelian. As far as I know, no similar classification exists for the $p=2$ case.
I think that a ``proper" question which must be proposed instead of Question 1, is the following:
Is there a classification of finite $p$-groups $G$ such that both $G'$ and $Z(G)$ are cyclic?
The latter question is inspired by the example of extra-special groups as in this case both center and commutator subgroups are cyclic with further conditions.
The answer of the above question is yes: To start finding a series of papers answering the question you may see:
A. A. Finogenov, Finite $p$-Groups With Cyclic Commutator Subgroup and Cyclic Center, Matematicheskie Zametki, Vol. 63, No. 6, pp. 911-922, June, 1998.
Classification of $p$-groups by nilpotency class is hard in general. As pointed above, coclass seems to be a better invariant. On the other hand, Ahmad, Magidin and Morse recently finished a classification of 2-generator $p$-groups of class 2:
Ahmad, Azhana(MAL-USM); Magidin, Arturo(1-LA); Morse, Robert Fitzgerald(1-EVAN-ECS) Two generator p-groups of nilpotency class 2 and their conjugacy classes. (English summary) Publ. Math. Debrecen 81 (2012), no. 1-2, 145–166.
(Addition to Isaacs' example) Let $G$ be a $p$-group and $d$ the minimal degree of representation of $G$ by permutations. Let $d<=p^n$. where $n$ is as small as possible. Then $G$ is a subgroup of $\Sigma_{p^n}$, a Sylow $p$-subgroup of the symmetric group of degree $p^n$. Since the center of $\Sigma_{p^n}$ has order $p$, we conclude that $G$ is a subgroup of a $p$-group with center of order $p$.
It is easy to prove that if a nonabelian $p$-group $G$ of exponent $>p$ contains $<p$ maximal abelian subgroups of exponent $>p$, then the Hughes subgroup of $G$ is abelian of index $p$.
PROBLEM. Classify the nonabelian $p$-groups of exponent $>p$ containing exactly $p$ maximal abelian subgroups of exponent $>p$.
If $G$ satisfies the problem, then $\exp(G)=p^2$ (Z. Janko). Moreover, all such $G$ are classified for $p=2$.
