An element $a$ of a ring is called idempotent if $a^2 = a$. In any ring, $0$ is idempotent, and $1$ is idempotent if it exists.
Let $D$ be an integral domain and let $a \in D$. Assume that $a^2 = a$. Then $a(a-1) = a^2-a = 0$, so $a = 0$ or $a-1 = 0$ (and in the latter case $a = 1$.)
Alternatively, note that if $a^2 = a$ then $a \cdot a = a \cdot 1$, so if $a \ne 0$ then we have $a = 1$ by cancellation.
In $\mathbb{Z}_{15}$ we have $0^2 = 0$, $1^2 = 1$, $2^2 = 4$, $3^2 = 9$, $4^2 = 1$, $5^2 = 10$, $6^2 = 6$, $7^2 = 4$, $8^2 = 4$, $9^2 = 6$, $10^2 = 10$, $11^2 = 1$, $12^2 = 9$, $13^2 = 4$, and $14^2 = 1$, so the idempotent elements are $0$, $1$, $6$, and $10$.
Under this isomorphism the idempotent elements $0$, $1$, $6$, and $10$ of $\mathbb{Z}_{15}$ map to the elements $(0,0)$, $(1,1)$, $(0,1)$, and $(1,0)$ of $\mathbb{Z}_3 \times \mathbb{Z}_5$ respectively, which are the idempotent elements of $\mathbb{Z}_3 \times \mathbb{Z}_5$.
We prove the contrapositive statement. Assume that $R'$ is isomorphic to $R$ and $R'$ is not an integral domain. We want to show that $R$ is not an integral domain. Take an isomorphism $\phi : R' \to R$ and take $a,b \in R'$ such that $ab = 0_{R'}$ and $a,b \ne 0_{R'}$. Then $\phi(a), \phi(b) \ne 0_R$ because $\phi$ is an injective homomorphism, and $\phi(a)\phi(b) = \phi(ab) = \phi(0_{R'}) = 0_{R}$, so $R$ is not an integral domain.
The function $\phi : \mathbb{Z} \to \mathbb{Z}_4$ defined by $\phi(a) = a \bmod 4$ is a surjective homomorphism, but $\mathbb{Z}$ is an integral domain and $\mathbb{Z}_4$ is not (because $2 \cdot 2 = 0$ in $\mathbb{Z}_4$.)
The function $\phi : \mathbb{Z}_4 \to \mathbb{Z}_2$ defined by $\phi(a) = a \bmod 2$ is a surjective homomorphism (because $2$ divides $4$,) but $\mathbb{Z}_4$ is not an integral domain and $\mathbb{Z}_2$ is.
Alternatively, the function $\phi : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $\phi((a,b)) = a$ is a surjective homomorphism, but $\mathbb{Z} \times \mathbb{Z}$ is not an integral domain and $\mathbb{Z}$ is.
(To see this, put the augmented matrix $(A|I)$ in reduced row echelon form. If the result is $(I|B)$ for some matrix $B$, then $B$ is an inverse for $A$. If not, then the equation $A \vec{x} = \vec{0}$ has at least one free variable.)
If the equation $A \vec{x} = \vec{0}$ has a nonzero solution $\vec{x} \in \mathbb{R}^n$, define the $n \times n$ matrix $B = (\vec{x}\;\vec{0}\;\cdots\;\vec{0})$. We have $B \ne O$ but $AB = (A\vec{x}\;A\vec{0}\;\cdots\;A\vec{0}) = O$, so $A$ is a zero divisor.
Let $R$ be a commutative ring with unity $1 \ne 0$ and let $a \in R$. Define a function $\ell_a : R \to R$ by $\ell_a(b) = ab$.
Assume that $\ell_a$ is injective. Then for all $b \in R$ such that $b \ne 0$ we have $\ell_a(b) \ne \ell_a(0)$, so $ab \ne 0$. Therefore $a$ is nonzero (consider $b = 1$ to see this) and $a$ is not a zero divisor.
Conversely, assume that $a$ is nonzero and $a$ is not a zero divisor. We want to show that $\ell_a$ is injective. Let $b_1,b_2 \in R$ and assume that $\ell_a(b_1) = \ell_a(b_2)$, which means that $ab_1 = ab_2$. So $a(b_1-b_2) = ab_1 - ab_2 = 0$. Then by our hypothesis on $a$ we have $b_1-b_2 = 0$, so $b_1 = b_2$.
Assume that $\ell_a$ is surjective. Then there is an element $b \in R$ such that $ab = \ell_a(b) = 1$. Because $R$ is commutative, $ba = 1$ also. So $a$ is a unit.
Conversely, assume that $a$ is a unit. We want to show that $\ell_a$ is surjective. Let $c \in R$. Then $\ell_a(a^{-1}c) = aa^{-1}c = c$.
Let $R$ be a ring with unity $1_R$ and let $S$ be a subring of $R$ with unity $1_S$.
If $1_R \in S$ then $1_R$ satisfies the definition of "unity of $S$" so $S$ has unity $1_S = 1_R$. Denote this common unity of $S$ and $R$ by $1$, and denote the common zero element of $S$ and $R$ by $0$. Then the characteristic of $S$ and the characteristic of $R$ are both equal to the least $n \in \mathbb{Z}^+$ such that $\underbrace{1+ \cdots + 1}_n = 0$, if such an $n$ exists, and zero otherwise.
Define $R = \mathbb{Z}_2 \times \mathbb{Z}_3$ and $S = \mathbb{Z}_2 \times \{0\}$. Then $1_R = (1,1) \notin S$, the characteristic of $S$ is $2$, and the characteristic of $R$ is $6$.
The ring $R = \mathbb{Z}_2 \times \mathbb{Z}_2$ is a counterexample. Its characteristic is $2$ (which is prime) because $1_R = (1,1)$ and $(1,1) \ne 0_R$ but $(1,1) + (1,1) = 0_R$. But $R$ is not an integral domain, because $(1,0) \cdot (0,1) = 0_R$.