The sequence (1+1/n)^n can be seen as a natural thing to study (indeed, tautologically, because we know that e is a natural thing to study) by various viewpoints.

The important aspect however is that the compound interest interpretation of this sequence is not only unnatural but wrong, in the sense that it starts from the wrong guess that a rate r compounded n times should be something like a rate r/n. It is unnecessarily confusing to teach such young students historical mistakes: bad students won't understand the mistake and good students will be bewildered. In both cases students come away thinking of e as a sort of "mystical" thing.

If I had to teach the compound interest, one starts by considering with a yearly interest rate r, your principal grows after m years by (1+r)^m. Now, can we find an equivalent monthly interest rate r2? To do so, we must solve (1+r2)^12 = (1+r), which requires logarithms, at which point you note immediately that r2 =/= r/12, which is perhaps unexpected. Now, it might be natural to ask what about for continuous compounding? Then, we study (1+f(n))^n as n goes to infinity, where f(n) is some function of n. We know it should decrease for large n, and the binomial expansion lets us guess (for integer n) that it should be f(n) = c/n for some constant c. Now, we are ready to compute the value of this limit.

The important part is that in studying compound interest one needs certain analytical tools, such as logarithms and asymptotic analysis, beforehand. It does not make much sense, as you say, to skip over all the intermediate steps and introduce the constant e as the solution to some unmotivated and unnatural formula relating to compound interest, but this is indeed what American schools do. In my experience, very few American high school students understand or remember (if they were ever taught it) the identity a^x = e^(x * ln a), and the concept of exponentiation is generally not well understood.

> To do so, we must solve (1+r2)^12 = (1+r), which requires logarithms

No, it doesn't. r_2 = (1 + r)^{1/12} - 1. Compound interest looks like (money) = (money_0)*r^t; you'd only need logarithms if you were trying to solve for time.

> Now, it might be natural to ask what about for continuous compounding?

I can't tell what you're getting at here. Once you've written down the equation (1 + r_2)^{12} = (1 + r), you've already provided a complete solution for continuous compounding. If the time you want to compound over is t, and Y is one year, then the solution is always given by (1 + r_2) = (1 + r)^{t/Y}. Nothing goes to infinity.