I have read the stated as a property with no proof. However, here in Konrad's article there is a mention, but in context of a different set of equalities:
$a \equiv b_1\pmod {m_1}, a \equiv b_2\pmod {m_2}$, i.e. $b_1, b_2$ are different quantities.
Also, the article differs by stating a necessary condition as $(m_1,m_2)=1,$ which is not stated in the text from which I picked it.
Also, my earlier post has failed to produce any proof for it.
I will answer the question: for given $m_1$ and $m_2$, find the conditions for $L$ such that the implication $a\equiv b\pmod {m_1}\land a\equiv b\pmod {m_2}\implies a\equiv b\pmod L$ is valid for all $a,b\in\mathbb Z$. If that is not what you mean, please shout.
Claim: The condition on $L$ is: $L\mid\operatorname{lcm}(m_1,m_2)$
Proof: $a\equiv b\pmod m$ is equivalent to $m\mid(b-a)$, so if you denote $d=b-a$, you are actually asking for conditions on $L$ such that $m_1\mid d$ and $m_2\mid d$ implies $L\mid d$.
Now, $m_1\mid d\land m_2\mid d\Longleftrightarrow \operatorname{lcm}(m_1,m_2)\mid d$, which means that the condition you are looking for is $L\mid\operatorname{lcm}(m_1,m_2)$.
Note: The proof of the equivalence $m_1\mid d\land m_2\mid d\Longleftrightarrow \operatorname{lcm}(m_1,m_2)\mid d$ depends on what you already know about $\operatorname{lcm}$. Actually, one definition of $\operatorname{lcm}$ is precisely that it is the minimal common multiple of $m_1$ and $m_2$, i.e. (i) It is a multiple (which proves $\Leftarrow$) and (ii) It divides every other multiple (which proves $\Rightarrow$). Depending how $\operatorname{lcm}$ is defined in your book, your mileage may vary.
Updated to add: After a discussion in the comments, it has transpired that, apparently, the question here is just to prove that $a\equiv b\pmod{m_1}$ and $a\equiv b\pmod{m_2}$ implies $a\equiv b\pmod L$ where $L=\operatorname{lcm}(m_1, m_2)$. The above proof still works, because $L=\operatorname{lcm}(m_1, m_2)$ is one of the numbers that divide $\operatorname{lcm}(m_1, m_2)$.