To limit my liability, let's only consider integers and then once we start talking about division, rational numbers. First off we intuitively understand addition:
If I have two apples and you give me three more apples, then I now
have 2 + 3 = 5 apples.
There are two routes we can go down: 1) we can accept negative numbers exist and deal with it or 2) we can accept subtraction as a valid operation that we definitely understand. I think the second approach is best for your question (since we do not want to accept negative numbers a priori).
So just as addition is intuitive to us, subtraction also is:
If I have 5 apples and you take 3 from me, then I am left with 2 apples: 5 - 3 = 2.
The value zero now becomes very important because I know that if I have $x$ apples and you take away $x$ apples then I am left with none, $0$:
$$
x - x = 0
$$
So now what happens when I have $5$ apples and you take away $6$? How many apples am I left with? Obviously, intuition now breaks down because you do not have $6$ apples to give up, but the math can remain:
$$
5 - 6 = 5 - (5 + 1) = 5 - 5 - 1 = 0 - 1
$$
We are happy in every step up until the last when I get to $0 - 1$ which we have no value for! I don't understand what $0 - 1$ represents in exactly the same way that I do not understand what $\sqrt{-1} = i$ represents--it's a definition! I am now defining that $0 - 1 = -1$--$-1$ is now a symbol for that value (which I do not fully comprehend). (and ultimately when I say $-x$, I really mean $0 - x$)
So now that we have this new symbol, what can we do with it? Well we can try and add it to values: $5 + -1 = 5 + (0 - 1) = 5 + 0 - 1 = 5 - 1 = 4$--we see that $5 + -1$ is the same as $5 - 1$! What about $5 - -1$? This is a little trickier. Now obviously we can write $5 - (0 - 1)$, but this doesn't help us because we don't know how to subtract a negative (in fact the above expression just devolves into $5 - -1$--the original question)! What we really need to show now is the following:
$$
0 - (0 - 1) = 0 - -1 = +1
$$
So we can do this through some algebraic manipulation:
$$
0 - (0 - 1) = x \\
0 = x + (0 - 1) \\
0 = x + 0 - 1 = x - 1 \\
0 + 1 = x + 1 - 1 = x + 0 = x\\
x = 1
$$
So notice that I used only addition to get to this result! This proves that $0 - -1 = +1$ therefore we can rewrite:
$$
5 - -1 = 5 + 0 - -1 = 5 + (0 - -1) = 5 + 1 = 6
$$
At this point, I hope that we both accept negative numbers as they are. The next question is for multiplication and division. If I have $5*-2$, then what should the result be? Well that one is easy:
$$
5*-2 = (-2) + (-2) + (-2) + (-2) + (-2) = -10
$$
What's not so easy is $-2*5$! There are two ways to approach this: 1) we accept that multiplication is commutative and thus $-2*5 = 5*-2 = -10$ (as we already showed) or 2) a negative multiplier means something "different" from a positive multiplier. A positive multiplier means to add the thing being multiplied whereas a negative multiplier means to subtract the thing being multiplied. The latter definition will help us define also a negative times a negative.
So what exactly is multiplication? Multiplication means taking a value and adding it to zero $x$ times (whatever the multiplier is). If the multiplier is negative, then it means subtracting from zero. For instance:
$$
5*-2 = 0 + (-2) + (-2) + (-2) + (-2) + (-2) = -10 \\
-2*5 = 0 - (5) - (5) = -10 \\
-2*-5 = 0 - (-5) - (-5) = 5 + 5 = +10 \\
-5*-2 = 0 - (-2) - (-2) - (-2) - (-2) - (-2) = +10
$$
From the above definition we see that a negative times a positive results in a negative value, a positive times a positive results in a positive value, and a negative times a negative results in a positive value. I don't want to go much further--division can be considered somewhat elementary (just as subtraction to addition) but, at this point, I think it's easier to accept division as the inverse of multiplication and prove the same laws apply (i.e. a division by a positive and negative gives a negative, etc.).