This is false for the cyclic group of order $6$. Let $g$ be a generator. Then $g^2-g+1$ acts by $0$ on the representations where $g$ acts by a primitive $6$-th root of $1$, and hence is not a unit in the group ring, but $1-1+1=1$ is relatively prime to $6$.

Generalizing this example, the statement is false for the cyclic group of order $pq$, with $p$ and $q$ two different primes. Let $g$ generate this group and let $\chi: G \to \mathbb{C}^{\ast}$ be a character with $\chi(g)$ a primitive $pq$-th root of unity.  Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\Phi_{pq}(g)= \sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\Phi_{pq}(\chi(g)) =0$ on the representation $\chi$. Thus $\sum c_k g^k$ is not a unit. On the other hand, $\sum c_k = \Phi_{pq}(1)= 1$. (To compute the last, note that $\Phi_{pq}(x) = \frac{(x^{pq}-1)(x-1)}{(x^q-1)(x^p-1)}$ and take the limit as $x \to 1$.)

I claim further that, if $G$ has any element of non-prime-power order, then $G$ fails to have this condition. Let $g$ be an element of order $pq$ and let $\chi: \langle g \rangle \to \mathbb{C}^{\ast}$ be an injective character. Let $V = \mathrm{Ind}_{\langle g \rangle}^G \chi$. Then $V$ restricted to $\langle g \rangle$ has $\chi$ as a summand, and this summand is in the kernel of $\Phi_{pq}(g)$ acting on $V$. So $\Phi_{pq}(g)$ acts non-injectively on a representation of $G$, and thus is not a unit.

So the only groups for which this might be right are groups where every element has prime power order. These were classified by [Higman][1], so you can dig into his paper if you care enough.

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On the positive side, the statement is true whenever $G$ is a $p$-group. Let $\alpha = \sum c_g g \in \mathbb{Z}[G]$. I will show that the determinant of $\alpha$ acting on $\mathbb{Z}[G]$ is $\left( \sum c_g \right)^{|G|} \bmod p$, and hence is not $0$ if $\sum c_g \not \equiv 0 \bmod p$. Reducing $\mathbb{Z}[G]$ modulo $p$, we get an action of $\alpha$ on $\mathbb{F}_p[G]$. More generally, I claim that $\alpha$ acts on any $G$-representation $V$ over $\mathbb{F}_p$ by $\left( \sum c_g \right)^{\dim V}$. This is simple: $V$ has a filtration whose associated graded is a $\dim V$-dimensional trivial representation. Passing to the associated graded doesn't change determinant, and $\alpha$ acts on the $1$-dimensional trivial representation by $\sum c_g$.


  [1]: https://doi.org/10.1112/jlms/s1-32.3.335