HINT: You can’t in general show that $\langle a,b\rangle$ belongs to $B$ or $C$; here’s a counterexample.
Let $L=\Bbb Z$, $m\land n=\gcd(m,n)$, and $m\lor n=\operatorname{lcm}(m,n)$. Let $A$ be congruence mod $2$, $B$ be congruence mod $12$, and $C$ be congruence mod $18$. Then $B\cup C$ is congruence mod $6$, so $\langle 2,8\rangle\in A\land(B\lor C)$, but $\langle 2,8\rangle$ is not in $A\land B$ or in $A\land C$.
For $a,b,c\in L$ define
$$M(a,b,c)=(a\lor b)\land(a\lor c)\land(b\lor c)\;,$$
and note that
$$M(a,a,b)=M(a,b,a)=M(b,a,a)=a$$
for any $a,b\in L$. The existence of such a function $M$ definable within $L$ is sufficient to ensure that $\operatorname{Con}\mathcal{L}$ is distributive.
Suppose that $\langle a,b\rangle\in A\land(B\lor C)$. Use (or prove and then use) the fact that $B\lor C$ is the transitive closure of $B\cup C$ to get $x_0,\ldots,x_{2n}\in L$ such that $x_0=a$, $x_{2n}=b$, and for $k=0,1,\ldots,n-1$ we have $\langle x_{2k},x_{2k+1}\rangle\in B$ and $\langle x_{2k+1},x_{2k+2}\rangle\in C$.
Verify that $\big\langle M(a,x_k,b),M(a,x_k,a)\big\rangle\in A$ for $k=0,\ldots,2n$. Use the fact that any congruence on $L$ must respect the function $M$.
Show that $$\big\langle M(a,x_{2k},b),M(a,x_{2k+1},b)\big\rangle\in A\land B$$ and $$\big\langle a,x_{2k+1},b),M(a,x_{2k+2},b)\big\rangle\in A\land C$$ for $k=0,1,\ldots,n-1$.
Conclude that $\langle a,b\rangle\in(A\land B)\lor(A\land C)$.
