Generalisation of integers for infinite length?

You've rediscovered the $10$-adic integers. Such "infinite decimal expansions of integers" form a ring using the usual rules for adding and multiplying decimals. However, they do not form a field. A quick and easy way to see this is that $10$ has no inverse (since if you multiply $10$ by anything, then the last digit of the product will be $0$).

Even worse, however, this ring has zero divisors: there are nonzero elements $x$ and $y$ such that $xy=0$. These take a bit of work to construct, but here's the idea. We'll construct the digits of $x$ and $y$ one at a time, starting from the end. Start by saying the last digit of $x$ is $2$ and the last digit of $y$ is $5$, so the last digit of $xy$ is $0$. Then choose the preceding digit of $x$ so that $x$ is divisible by $4$, and the the preceding digit of $y$ so that $y$ is divisible by $25$, so their product will be divisible by $100$ and end in two zeroes (for instance, $x$ might end in $12$ and $y$ might end in $25$). Continue choosing digits of $x$ and $y$ one at a time so that $x$ is divisible by every power of $2$ and $y$ is divisible by every power of $5$. Every digit of the product $xy$ will end up being $0$.

You can, of course, do the same thing with a different base instead of $10$, giving you the $b$-adic integers for any integer $b>1$. Unlike the case of finite base expansions, however, you get a genuinely different number system for different values of $b$! (To be precise, it turns out that the number system you get depends only on the set of prime factors of $b$, so for instance the $10$-adic integers are isomorphic to the $50$-adic integers.)

It turns out that whenever $b$ is composite, the $b$-adic integers will have zero divisors, as in the case $b=10$. When $b$ is a prime (say $b=p$), however, the $p$-adic integers are an integral domain (they have no zero divisors). The $p$-adic integers aren't quite a field, though, because $p$ has no inverse. If you adjoin an inverse to $p$, you do get a field, called the $p$-adic numbers and written $\mathbb{Q}_p$. Elements of $\mathbb{Q}_p$ are "base $p$ expansions that can be infinite on the left": that is, formal sums $$\sum_{n=k}^\infty a_np^n$$ where each $a_n$ is an integer between $0$ and $p-1$ and $k\in\mathbb{Z}$ (so $k$ can be negative, allowing finitely many negative powers of $p$ in the sum).

There is a much larger story here: $p$-adic numbers play a huge role in modern number theory, and have a very rich structure. Since it is easily possible to write an entire book on the topic, I'll end my answer here though. You can read a bit more about them on the Wikipedia page I linked at the beginning.

Yes. What you've defined is called the 10-adic numbers. They don't form a field for two reasons, one of which is that $10$ is not invertible; the other reason is left as an exercise.

Most books will only talk about the $p$-adic numbers for $p$ prime, and this is because the general case turns out to reduce to the prime case, essentially because of the Chinese remainder theorem.

Another fun way that the rings differ depending on the base used is that in base 5, $-1$ (i.e. $\bar{4}$) has a square root while $2$ doesn't, but in base 7 it's the other way around (and $-1$ is of course $\bar{6}$ in that base).

I should note that I haven't proved those, but my messing around with the idea pointed strongly that way. If you like programming it's a fun exercise to write up a class that implements these numbers. I did it in Python, as its yield/generator system made it easy to model these numbers as a potentially infinite list that produced the digits from units place up.