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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. Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\sum c_k \chi(g)^k = \Phi_{pq}(\omega) =0$ 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, so you can dig into his paper if you care enough.

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 $\sum c_g$.

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, so you can dig into his paper if you care enough.

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 $\sum c_g$.

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 $6$.

More generally, the statement is false whenever $G$ has an abelian quotient not of prime power order. Let $p$ and $q$ be two different primes dividing $G^{\mathrm{ab}}$, so there is a surjection $G \to \mathbb{Z}/(pq)$. Let $\chi$ be the character of $G$ formed by composing $G \to \mathbb{Z}/(pq)$ with an injection $\mathbb{Z}/(pq) \to \mathbb{C}^{\ast}$. Let $g \in G$ be such that $\chi(g)$ is a primitive $pq$-th root of unity, $\omega$.

Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\sum c_k \chi(g)^k = \Phi_{pq}(\omega) =0$ 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 $x \to 1$.)

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 $\sum c_g$.

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. Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\sum c_k \chi(g)^k = \Phi_{pq}(\omega) =0$ 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, so you can dig into his paper if you care enough.

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 $\sum c_g$.

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 $6$.

More generally, the statement is false whenever $G$ is a solvable group not of prime power order. Let $p$ and $q$ be two different primes dividing $|G|$. Then $p$ and $q$ also divide $G^{\mathrm{ab}}$, so there is a surjection $G \to \mathbb{Z}/(pq)$. Let $\chi$ be the character of $G$ formed by composing $G \to \mathbb{Z}/(pq)$ with an injection $\mathbb{Z}/(pq) \to \mathbb{C}^{\ast}$. Let $g \in G$ be such that $\chi(g)$ is a primitive $pq$-th root of unity, $\omega$.

Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\sum c_k \chi(g)^k = \Phi_{pq}(\omega) =0$ 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 $x \to 1$.)

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 $\sum c_g$.

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 $6$.

More generally, the statement is false whenever $G$ has an abelian quotient not of prime power order. Let $p$ and $q$ be two different primes dividing $G^{\mathrm{ab}}$, so there is a surjection $G \to \mathbb{Z}/(pq)$. Let $\chi$ be the character of $G$ formed by composing $G \to \mathbb{Z}/(pq)$ with an injection $\mathbb{Z}/(pq) \to \mathbb{C}^{\ast}$. Let $g \in G$ be such that $\chi(g)$ is a primitive $pq$-th root of unity, $\omega$.

Let $\Phi_{pq}(x) = \sum c_k x^k$ be the $pq$-th cyclotomic polynomial. So the element $\sum c_k g^k$ in $\mathbb{Z}[G]$ acts by $\sum c_k \chi(g)^k = \Phi_{pq}(\omega) =0$ 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 $x \to 1$.)

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 $\sum c_g$.