By restricting it to pairs that are number bonds to 10, the only pairs allowed are:

1+9, 2+8, 3+7, 4+6, 5+5

plus the 10s on their own. The 10s on their own are a bit of a red herring as they take care of themselves i.e. they can always be removed without needing to find a partner card with the appropriate face value.

You mention two rows of five, is there a requirement that when you remove a pair, one card be drawn from one row and the other card from the other row? Assuming not…

There can't be any mathematical proof that this will always work out, because there are simple counterexamples, e.g. if the first palette of 10 randomly drawn cards have values 1, 1, 1, 1, 2, 2, 2, 3, 3. (You could randomly draw four aces, four 2's and two 3's and be unable even to get started on the game).

However, so long as you’re never unlucky enough to end up with a stubborn palette like this at any stage, the game will always work out. There will always be four individual 10s to extract, two 5+5 pairs (suite A’s 5 + suite B’s 5, suite C’s 5 + suite D’s 5) and four of each of the other pairs (here they can be drawn from mixed and/or matching suites, but it makes no difference, there’s simply a pool of (e.g.) four 1’s and four 9’s which can be paired off in any order leaving no 1’s and no 9’s). So once the deck is empty, the (non-stubborn) palette will always work out, because you're just finishing up this pairing process -- but by restricting the visibility of the pairing process to a window of 10 cards, the illusion is given that this is magical.

The exact probability of ever getting a stubborn palette would be very difficult to work out because of the complex way the palette evolves by chance during the course of the game. But you could deal with this by working out the chance that any randomly drawn palette will contain at least one good pair, and then showing that the evolving palette during the game is always approximately as good for finding pairs as a random palette – i.e. that given a random palette, the act of removing a pair or a 10 and drawing new cards doesn’t mean you’ll be significantly worse off that with a new random palette from a fresh pack.

To work out the chance with a random palette, you may as well ignore the 10’s, because the game player could simply go on substituting them out for cards from the deck until there are no 10’s in the palette. We can calculate a worse-case bound on the chance by assuming there is exactly one 5 (maximum stubbornness) and looking at there being nine further cards in the palette, drawn from the set:

1111222233334444   6666777788889999

So the simplified question is, if you draw 9 cards from the above set, what is the chance that all 9 come from the same half. (If so, there will be zero pairs bonding to 10).

Well, the first card can only come from one half. For the second card, the chance of it being from the same half is 15/31 (there are 31 cards left, and 15 of them are in the same half as the half whose 16^th card was already removed). This goes on:

15/31  (2^nd card)

14/30  (3^rd card)

13/29

12/28

…

8/24 (9^th card)

Multiplying these (because they ALL have to happen for all 9 cards to be from the same half) yields an overall probability of around 0.0008, or 0.08%. A random configuration has a less than 1 in 1,000 chance of containing no valid pairs: counterexamples like 1, 1, 1, 1, 2, 2, 2, 3, 3 are very unlikely to occur in practice.

So how about this question of whether the chance could grow significantly as you remove the pairs over the course of the game. Well, removing a pair doesn’t significantly skew the “balance of halves” (i.e. the proportion of cards in the palette that are from the 1-4 half versus the 6-9 half) because removing the pair always removes one from each half. Then we’re just adding two further random cards from the deck. And if the balance of halves in the palette is fairly even, the balance of halves in the deck will also be fairly even, so the chance they will be from the same half is similar to the chance from the full deck. And if the balance in the palette is a bit uneven, the balance in the deck will be uneven in the opposite direction  (e.g. if your palette has mostly 1-4's, your deck will have mostly 6-9's -- bearing in mind that the pile of removed pairs will always be perfectly evenly balanced) so the drawn cards will be tend to reduce the imbalance. Any imbalance has a tendency to right itself over the course of the game.

This skips over some awkward details, such as removal of 10's and 5+5's, but these don't make any significant difference -- overall, the balance of a derived palette is likely to be similar to the balance of the initial palette, and the chance that the balance of an initial (random) palette is stubborn is extremely low. (In a similar way, the probability that it's significantly skewed can also be shown to be low). So it’s extremely unlikely that the game will ever not work out.