It is possible to flip a single card in one direction without affecting any other cards, like this:
A B C B C a D C a C a d B a d a d b C d b D B C a B C
D E F D E F G E F G E F C E F C E F G E F G E F D E F
G H I G H I b H I b H I G H I G H I A H I A H I G H I
L U L D L U RR D
Here only moves of the top row and left column are used to flip the card in the top left corner horizontally.
By applying the same moves to different rows/columns you can flip any card, and by applying the moves 90 degrees rotated you can flip any card along the other axis.
This means that for any permutation of the cards, all $4^9$ orientations of those cards is possible.
However, not all $9!$ permutations are possible. Every move is a 3-cycle, which is a permutation with even parity. Combining even permutations only results in even permutations, so no odd permutations can be achieved. It is fairly easy to show that any even permutation is solvable using the 3-cycles you have available (for example by using the well-known theorem that the 3-cycles of the form $(1,2,k)$ generate all even permutations).
The total number of achievable states is therefore $4^9\cdot\frac{9!}2 = 47,563,407,360$, and a randomly chosen shuffled deck of cards has a $50\%$ probability of being solvable.
You can fix this problem by making two cards identical, for example by making cards 8 and 9 blank apart from their colours. By doing that, every shuffled deck becomes solvable because it becomes possible to swap any two numbers. This is because a swap of two numbered cards can be combined with a swap of the two identical blank cards, and such a double-swap is an even permutation and therefore achievable using the moves of the game. Note that the number of permutations is still $9!/2$, so having two unnumbered cards does not reduce the number of solvable positions of the game.
Edit:
For even grid sizes you should still use two blank cards, but it is for a slightly more complicated reason.
If you only look at the permutations of the numbers when both dimensions of the grid are even, then then every permutation can be reached/solved. So you'd think that no blank cards are needed in that case. However, something interesting happens with the orientations when the dimensions are even. Every move changes the parity of the permutation, but also does exactly one card flip. So the total number of card flips will be even or odd, and that must match the parity of the permutation. Both parities are affected by every move, they are equal (both even) in the solved state, so they will remain equal to each other throughout the game and must be equal for the position to be solvable.
Having two cards without numbers will still ensure that every shuffled deck will lead to a solvable position, without reducing the number of positions that can be played.
Something similar happens if only one of the dimensions is odd, but then it is not the total number of flips but the number of flips in the even direction that has to match parity with the permutation. Flips along the odd direction can be done in essentially the same way as in the 3x3 without changing the permutation. So again, using two unnumbered cards will allow the use of a shuffled deck.
In this last case with an even by odd sized grid, then there is a different way to fix the shuffled deck problem. Instead of two unnumbered cards, you could instead have one anomalous card which does not change when flipped over in the even direction. Normally it would be impossible to flip just a single card in the even direction without affecting the numbers (it would have to be an odd permutation of the numbers to match the odd card flip, so some numbers would have to be moved). It is possible however to combine the flip of that single card with a flip of the anomalous card. Flipping both cards can be done without affecting the number permutation, and since the flip of the anomalous card is not visible it is just as if a single card was flipped. Therefore the presence of this anomalous card makes every shuffled state solvable.
Best Answer
The entire first chapter of Engel's famed olympiad training book, Problem-Solving Strategies is on "The Invariance Principle". There are several illuminating examples followed by sixty problems and their solutions. The second chapter in the book is on "Coloring Proofs," a part of which is about parity, and that often also involves invariants (and showing that some configuration, say on a chessboard, is impossible because it violates the parity invariant).
I recommend also looking into monovariants which change in some predictable way, unlike invariants which stay constant; monovariants are sometimes mistakenly grouped under invariants.
Pranav Sriram's free online book on Olympiad Combinatorics contains material on advanced olympiad problem solving using invariants and monovariants. It takes some effort to find a PDF that contains all the chapters though; I recall that there are nine chapters but they were posted separately at different times.
Looking up "invariant" in puzzling.SE yielded results so you could try that.