Let $G$ be a group, which is
Is it possible to choose generators such that one of the generators is central?
The two answers below give counter-examples, when $G$ is only assumed to satisfy (1) and (3) or (1), (2), and (3).
On
Take a non abelian group of order $\,|G|=p^3\,\;,\;p\;$ a prime, with exponent $\,p\,$. In this case, we have (with $\,\Phi(G)=$ the Frattini subgroup of $\,G\,$):
$$G'=Z(G)\;,\;\;\Phi(G)=G^pG'\implies Z(G)=G'\le\Phi(G)$$
and this means that every central element is a non-generator of $\,G\,$ , so the answer to your question is no.
My previous answer below was given when condition (4) was not there in OP.
With condition (4) added, the answer is yes. The solution I am giving is surely more complicated than it is necessary.
We know $G / Z(G)$ is torsion-free and abelian. Suppose by way of contradiction that $G/Z(G)$ has rank three.
Fix generators $a, b, c$ of $G$ so that $[a, b] \ne 1$, and a generator $d$ of $G'$, so that $G' = \langle d \rangle \le Z(G)$.
Consider the group homomorphism $G/Z(G) \to G'$ given $x \mapsto [x, a]$. If this maps onto $\{ 1 \}$, we are done.
Writing elements of $G$ as product of powers of $a, b, c$, and elements of $G'$ as powers of $d$, we obtain a group homomorphism $$ \alpha : \mathbf{Z}^{3} \to \mathbf{Z}. $$ To make things slightly clearer, extend this to a $\mathbf{Q}$-linear map $$ \alpha' : \mathbf{Q}^{3} \to \mathbf{Q}. $$ Do the same with $x \mapsto [x, b]$, obtaining $$ \beta : \mathbf{Z}^{3} \to \mathbf{Z}, \qquad \beta' : \mathbf{Q}^{3} \to \mathbf{Q}. $$
Now $\ker(\alpha')$ and $\ker(\beta')$ are subspaces of $\mathbf{Q}^{3}$ of dimension $2$, so they intersect in a one-dimensional subspace $\langle k \rangle$ of $\mathbf{Q}^{3}$ Clearly a multiple of $k$ will be in $\mathbf{Z}^{3}$, and this corresponds to an element $$ f = a^{s} b^{t} c^{u} \ne 1 $$ such that $[f, a] = [f, b] = 1$. In particular, $f \in Z(\langle a, b, f \rangle)$.
Since we have taken $[a, b] \ne 1$, we have $u \ne 0$. Then $c^{u} \in \langle a, b, f \rangle$, so $$ 1 = [c^{u}, f] = [c, f]^{u}, $$ which implies $[c, f] = 1$ because $G$ is torsion-free, so that $f$ is central. But then the normal subgroup $\langle a, b, Z(G) \rangle$ has finite index at most $u$ in $G$, and this contradicts the fact that $G/Z(G)$ has rank three.
Previous answer
No. A counterexample is the free group in the variety of groups of nilpotence class at most $2$ ($2$-step nilpotent for you).
To be explicit, start with the free abelian group of rank $4$ $$ H = \langle a_{1}, c_{12}, c_{13}, c_{23} \rangle \cong \mathbf{Z}^{4}, $$ then extend it by the automorphism $a_{2}$ of infinite order such that $$ a_{1}^{a_{2}} = a_{1} c_{12}, \qquad c_{ij}^{a_{1}} = c_{ij}, $$ and then extend $K = \langle a_{1}, a_{2}, c_{12}, c_{13}, c_{23} \rangle$ by the automorphism $a_{3}$ of infinite order such that $$ a_{1}^{a_{3}} = a_{1} c_{13}, \qquad a_{2}^{a_{3}} = a_{2} c_{23}, \qquad c_{ij}^{a_{1}} = c_{ij}. $$
In the resulting group $G = \langle a_{1}, a_{2}, a_{3} \rangle$ the derived subgroup is free abelian of rank $3$, $G' = \langle c_{12}, c_{13}, c_{23} \rangle \cong \mathbf{Z}^{3}$. (This is because $[a_{i}, a_{j}] = a_{i}^{-1} a_{j}^{-1} a_{i} {a_{j}} = a_{i}^{-1} a_{i}^{a_{j}} = c_{ij}$ for $i < j$.)
But if you take any subgroup of $G$ with three generators, one of which is central, its derived subgroup will be cyclic (of rank $1$), generated by the commutator of the two non-central generators.
For a simpler alternative, consider the abelian group of rank $4$ $$ L = \langle a_{2}, a_{3}, c_{21}, c_{31} \rangle \cong \mathbf{Z}^{4} $$ and extend it by the automorphism $a_{1}$ such that $$ a_{2}^{a_{1}} = a_{2} c_{21}, \qquad a_{3}^{a_{1}} = a_{3} c_{31}, \qquad c_{ij}^{a_{1}} = c_{ij}. $$ The resulting group $M = \langle a_{1}, a_{2}, a_{3} \rangle$ has derived subgroup $M'= \langle c_{21}, c_{31} \rangle$ of rank $2$, so the same argument applies.