Let $n = lcm(a, b)$, where $a$ is the first ideal and $b$ is the second ideal. So for any $(p, q) \in A$ suppose $n(p,q) = 0$, so we have $np = 0\mod a, nq = 0 \mod b$. Thus $a | np$ and $b | nq$. Thus we have $lcm(p, q) = 0$, which proves the $lcm(a, b)$ is the characteristic of the cartesian product of the two rings.

Is that correct?

Let $R$ and $S$ be rings with (multiplicative) identity. What are the units in $R×S$?

Obviously those $(r_1, s_1) \times (s_1, r_1) = (1, 1)$.

Am I leaving out any important details here?

"Let $A$ be an $m\times n$ matrix. Suppose that $A$ has a right inverse $B$ and has a left inverse $C$. Prove that $m=n$, $B=C$, and $B$ is the inverse of $A$."

Proof:
Since $A$ has a right inverse $B$, we can conclude that $rank(A)=m$. Aditionally, since $A$ has a left inverse $C$, we can conclude that $rank(A)=n$. Therefore, $m=n$, and $A$, $B$, and $C$ are square matrices. Since $m=n$, we have $CA=I=AB$, $C=B=A^{-1}$.

Am I leaving out any important details here?

"Let $A$ be an $m\times n$ matrix. Suppose that $A$ has a right inverse $B$ and has a left inverse $C$. Prove that $m=n$, $B=C$, and $B$ is the inverse of $A$."

Proof:
Since $A$ has a right inverse $B$, we can conclude that $rank(A)=m$. Aditionally, since $A$ has a left inverse $C$, we can conclude that $rank(A)=n$. Therefore, $m=n$, and $A$, $B$, and $C$ are square matrices. Since $m=n$, we have $CA=I=AB$, $C=B=A^{-1}$.