Your expected gains (or rather losses) are the same for both methods. However, if you get tickets for separate draws, there is an ever so tiny chance that you will win more than once, and correspondingly the chance that you will win (at least once) will be an ever so tiny bit smaller.
As an extreme example of this phenomenon, replace $10$ by the total number of tickets in one draw. Then taking them all in the same lottery ensures a win in that lottery, but taking them in all different lotteries does not ensure any win, but might lead to multiple wins.
This is a tricky problem. The lottery people would love for you to think of the problem simplistically so you arrive at the wrong answer. However, a careful analysis shows why the lottery people really know why they will make tons of money from even a huge payout.
Let $p$ be the probability of winning, $C$ be the cost of a ticket, and $V$ be the value of the winnings. Then the expected value $E$ of a ticket, assuming one winner, would be approximately
$$E = p V - (1-p) C$$
However, this is really not correct. In reality, there will be more than one winner. Or none. Who knows? But when there are more than one winner, the value of the the winnings to each person are reduced, as the winnings are split evenly among the winners.
It may be assumed that the number of winners follows a binomial distribution. Assume a population of $N$ possible tickets. The probability of $k$ tickets, including yours, being winners is
$$\binom{N-1}{k-1} p^k (1-p)^{N-k} $$
where $k =1$ corresponds to the ideal case in which you alone are the winner, and k may vary between $1$ and $N-1$. so that the actual expected value of your winning ticket is equal to
$$E = \sum_{k=1}^{N} \binom{N-1}{k-1} p^{k} (1-p)^{N-k} \frac{V}{k} - (1-p) C $$
which may be simplified to
$$E = \frac{1-(1-p)^N}{N} V - (1-p) C $$
Note this takes into account the number of tickets purchased and will reduce the expected value of the ticket from the simpler assumption.
Given the numbers: $p$ being $1$ in $282.2$ million, $V=\$700$ million, and $C=\$2$, this distinction is crucial. The expected value for the simple case $N=1$ is positive (about $ \$0.396 $); people who understand the expected value at this level may be induced to buy a ticket, thinking that each ticket has positive value. However, one may show that, when there are more than about $108.8$ million tickets sold, the expected value goes negative. My guess is that the number of tickets sold will certainly exceed this number and that the lottery people will make a profit.
Best Answer
This is conditional expected probability. Given that you scratched a bunch of tickets, you know that the probability of the last one being a winner is increased only because you know that the tickets are not independently random; there is a set $1000$ out of a million winners.
However, if each ticket was independently random (i.e each ticket had a $0.1\%$ win rate) then you'd still have the same chance.
Back to your other paradox: given that you didn't win, you now know that all remaining tickets have an increased chance of being a winner, including those at the store. But that's only knowing that you didn't win - a random stranger couldn't use this logic without knowing that you didn't already win.
One last note, in real life this doesn't actually matter much as the number of tickets produced is so huge that even buying a thousand tickets would barely make a dent in the 'odds' of each one, and besides I'm not sure if lottery companies ensure there are at least X winners although I could be wrong.
Edit: answering your title question, no, the tickets are all equally valuable before you scratch yours. After you scratch yours, the ones at the store are actually slightly more likely of being winners, but there was no way to work that out without buying your tickets first.