Let $s,r$ be natural numbers such that $s \geq r$, $$B = \prod_{i = 1}^r\mathbb{Z}_{e_i}$$
a finite abelian group where $1 < e_r$ and $e_{i+1} | e_i$ for all $i$ and consider $\phi$ a surjective map $\phi: \mathbb{Z}^s \rightarrow B$.
Since the map is surjective, we can find an element $g_i \in \mathbb{Z}^s$ such that $\phi(g_i) = w_i = (0,...,1,...,0)$ where the $1$ is in the $i$-th coordinate. The question is:
can we for all $i$ find $g_i$ such that at least one coordinate is coprime to $e_1$?
Example: Take $\mathbb{Z}^2 \rightarrow \mathbb{Z}_6 \times \mathbb{Z}_2$ where $(x,y) \mapsto (2x+y,x+y)$. This map is surjective because $(3,0) \mapsto (0,1)$ and $(1,-1) \mapsto (1,0)$. Observe that $(3,0)$ does not satisfy my claim however $(2,2)$ is in the kernel and consequently $(5,2) = (3,0) + (2,2) \mapsto (0,1)$ and at least one coordinate is coprime to $e_1 = 6$.
Progress in the problem: If $e_1$ is a prime power $p^\alpha$ the result is true. In fact, suppose that $g_i$ has no coordinate coprime to $e_1$, then $g_i = pg_i'$ for some other vector $g_i' \in \mathbb{Z}^s$. Then looking at the $i$-th coordinate, $$1 = \phi(g_i)_i = \phi(pg_i')_i = p \phi(g_i') \pmod{e_i}$$ getting a contradiction since $p$ is not a unit modulo $e_i$.