Do the following equalities hold in any Heyting algebra(where $\land,\lor$ are the classical lattice operations):
$$ \bigvee_{i\in I}\{x_i\land y_i\} = \bigvee_{i\in I}\{x_i\}\land\bigvee_{i\in I}\{y_i\} $$
$$ \bigvee_{i\in I}\{x_i\lor y_i\} = \bigvee_{i\in I}\{x_i\}\lor\bigvee_{i\in I}\{y_i\} $$
$$ \bigvee_{i\in I}\{x_i\rightarrow y_i\} = \bigvee_{i\in I}\{x_i\}\rightarrow\bigvee_{i\in I}\{y_i\} $$
I'll assume that those joins exist (for example, in complete Heyting algebra).
Then concerning the first equality, it is not exactly true the way it is stated, even if $I$ is finite.
For example, if $I=\{1,2\}$, it becomes $$(x_1 \vee x_2) \wedge (y_1 \vee y_2) = (x_1 \wedge y_1) \vee (x_2 \wedge y_2),$$ when the truth is that $$(x_1 \vee x_2) \wedge (y_1 \vee y_2) = (x_1 \wedge y_1) \vee (x_1 \wedge y_2) \vee (x_2 \wedge y_1)\vee (x_2 \wedge y_2).$$ So I suppose what you really mean in the first equality is $$\bigvee_{i,j \in I} (x_i \wedge y_j) = \bigvee_{i \in I}x_i \wedge \bigvee_{i \in I}y_i.$$ Check the infinite distributive law for Heyting algebras. By applying it twice you get that equality.
Concerning the second identity, we have $\bigvee_{i \in I}(x_i \vee y_i) \geq \bigvee_{i \in I} x_i$ and $\bigvee_{i \in I}(x_i \vee y_i) \geq \bigvee_{i \in I} y_i$ simply because $x_i \vee y_i \geq x_i, y_i$. So it follows that $$\bigvee_{i \in I}(x_i \vee y_i) \geq \bigvee_{i \in I} x_i \vee \bigvee_{i \in I} y_i.$$ On the other hand, for each $i$, $x_i \vee y_i \leq \bigvee_{i \in I} x_i \vee \bigvee_{i \in I} y_i$ and so you get the desired equality.
The third one is false. Indeed, $$a \to b = \max\{c : a \wedge c \leq b\},$$ so that $a \leq a'$ implies $a'\to b \leq a \to b$, and so $\to$ cannot preserve joins in the first coordinate.
Edit. There is something else you may find useful about the relation between the lattice operations and the implication (as an alternative to the identity you proposed).
It is not difficult to show that $$\bigvee_{i \in I}x_i \to y = \bigwedge_{i \in I}(x_i \to y)$$ and $$x \to \bigwedge_{i \in I}y_i = \bigwedge_{i \in I}(x \to y_i)$$ and thus, $$\bigvee_{i \in I}x_i \to \bigwedge_{i \in I}y_i = \bigwedge_{i,j \in I}(x_i \to y_j).$$