Losslessness of Laplacian Pyramid
Actually the down sampling has no role here.
It is all based on a real simple equation:

$$ I = A + B $$

It is always enough to keep 2 terms of the 3 to restore completely and perfectly the information.

So let's look on this:

$$ {I}_{0} = \left( \left( {I}_{0} \downarrow \right) \uparrow \right) + {R}_{0} $$

So if we keep $ {R}_{0} $ and we have $ \left( \left( {I}_{0} \downarrow \right) \uparrow \right) $ we can recover $ {I}_{0} $ right?
But we can fully recover $ \left( \left( {I}_{0} \downarrow \right) \uparrow \right) $ form $ \left( {I}_{0} \downarrow \right) $ as it is deterministic known operation (Up sampling) right?

So instead of $ {I}_{0} $ we will keep $ {I}_{1} = \left( {I}_{0} \downarrow \right) $ and $ {R}_{0} $.  

We now can repeat the same trick for $ {I}_{1} $:

$$ {I}_{1} = \left( \left( {I}_{1} \downarrow \right) \uparrow \right) + {R}_{1} $$

So instead of keeping $ {I}_{1} $ we can again keep $ {I}_{2} = \left( {I}_{1} \downarrow \right) $ and $ {R}_{1} $.
So in order to resotre $ {I}_{1} $ perfectly all we need is $ {I}_{2} $, $ {R}_{1} $ and $ {R}_{0} $.

Keep doing this and you will get the Laplacian Pyramid which says keep the residual of each level and the Low Pass image of the last image.

The trick here isn't the accurate definition of $ \downarrow $ or $ \uparrow $ but being able to get the same result every time you use them. Namley:

$$ \left( \left( \left( \left( I \downarrow \right) \uparrow \right) \downarrow \right) \uparrow \right) = \left( \left( I \downarrow \right) \uparrow \right) $$