1. Whether the classical and non-standard definitions of continuity and uniform continuity are the same, rigorously speaking, or just extremely similar?
The classical and nonstandard definitions of continuity are the same, for real functions, in the sense that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the usual standard continuity condition precisely if its $\:^\star$-extension $\:^\star\! f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ satisfies the nonstandard continuity condition presented in the linked article.
Similarly, $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the usual standard uniform continuity condition precisely if its $\:^\star$-extension $\:^\star\! f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ satisfies the nonstandard uniform continuity condition presented in the linked article.
However, the usual definitions of continuity and microcontinuity do not coincide for arbitrary hyperreal functions $g : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$. This happens because most such functions are not $\:^\star$-extensions of any real function. Assuming that your usual standard definition of continuity and your definition of microcontinuity are both phrased in such a way that they make sense for hyperreal functions, you get examples of hyperreal functions that are continuous in the usual standard sense but not in the nonstandard sense (the indicator function of the non-real hyperreal numbers) and vice versa (the indicator function of the rationals multiplied by an infinitesimal).
I cannot imagine a counterexample of a real function which is classically continuous but not microcontinuous!
This one's easy to settle. Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by the equation $f(x) = x^2$. Take your favorite infinite hypernatural $\omega$. This hypernatural is positive, and larger than any natural number. Since if $x \leq y$ then $\frac{1}{y} \leq \frac{1}{x}$ for all $x$, $y$, we get that $\frac{1}{\omega}$ is a positive hyperreal that is smaller than any positive natural number, and is therefore infinitesimal.
So we have that $\omega$ is infinitesimally close to $\omega + \frac{1}{\omega}$. Is $f(\omega)$ infinitesimally close to $f\left(\omega + \frac{1}{\omega}\right)$? Well, $\left(\omega + \frac{1}{\omega}\right)^2 - \omega^2 = \omega^2 + 2\omega\frac{1}{\omega} + \frac{1}{\omega^2} - \omega^2 = 2 + \frac{1}{\omega^2} > 2$, which is therefore not infinitesimal. So $f(\omega)$ and $f\left(\omega + \frac{1}{\omega}\right)$ are not infinitesimally close, and consequently $f$ is not microcontinuous, since if you take $x=\omega$ and $x' = \omega + \frac{1}{\omega}$, then the premise of the implication is satisfied, but the conclusion is not. And this same argument shows that the real function $f$ is not uniformly continuous on the real line.
Which brings us to uniform continuity. You write:
This seems to align itself fairly well with the standard uniform continuity definition: $\forall x,y\in I,\,\forall\epsilon\gt0,\,\exists\delta\gt0:\|x-y\|\lt\delta\implies\|f(x)-f(y)\|\lt\epsilon$
This is not the definition of uniform continuity at all. You need $$\forall \varepsilon > 0. \exists \delta > 0. \forall x, y \in I. |x - y|<\delta \rightarrow |f(x)-f(y)| <\varepsilon$$
instead. Your reasoning, that "the counterexample of a continuous-but-not-microcontinuous function is a function that is not uniformly continuous anywhere" does not work. In fact, uniform continuity is not a local property: a continuous real function is uniformly continuous on every closed and bounded real interval, so it doesn't make much sense to talk about a continuous function that is not uniformly continuous anywhere. This has nothing to do with nonstandard analysis: it should be familiar from introductory courses on standard real analysis.
2. How can a function ever be continuous everywhere but not microcontinuous anywhere?
There is no such real function. If a real function is continuous, then it is microcontinuous around every real point. But possibly not around non-real hyperreal points (see the squaring function above).
There are such hyperreal functions, though (provided that you define microcontinuity appropriately for such functions). Consider e.g. the function $f : \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{R}$ that takes the value $0$ on every real number, but takes the value $1$ on every non-real hyperreal number. For any $x \in \mathbb{R}, \varepsilon \in \mathbb{R}^+$, there is clearly a $\delta \mathbb{R}^+$ so that for all $y \in \mathbb{R}$, if $|x - y| < \delta$ then $|f(x) - f(y)| = |0 - 0| = 0 < \delta$. So $f$ is continuous. But if you take $x = 0$ and any non-zero infinitesimal $x' \approx 0$, then $1 = f(x') \not\approx f(0) = 0$. So $f$ is not microcontinuous.
Whether or not there exists analogous definitions of most standard analysis ideas of convergence, continuity and limits (with all the typical extra statements of uniform, weak, strong, absolute,...) in the nonstandard world?
There are analogous definitions of all these. You can find them in any nonstandard analysis textbook, e.g. Goldblatt's Lectures on the Hyperreals.
Best Answer
The biggest draw back (and it's a big one) is that the ring of dual numbers is not a field. It has plenty of zero divisors. So, Newton, or any of the mathematicians of the early days of calculus, certainly did not work directly in the ring of dual numbers. They of course did not consider the ring to exist (as rings did not exist at all yet), but from their writing it is clear they envisaged a field of real numbers with, somehow, some notions of infinitesimals. Their work is of course very vague, but correct. Much more on that can be found in math history books. Many interesting discussions can be found in the recent book "Adventures in Formalism", also related to the early days of calculus and how things developed.
Some (rather unsatisfactory) portions of analysis can be developed in the ring of dual numbers, but it does not go too far. The idea, as you say, is very simple, perhaps too simple. One immediately gets into trouble when trying to define the derivative as the quotient of the infinitesimal $f(x+h)-f(x)$ divided by $h$, where $h$ is infinitesimal. The difficulty is that the non-zero infinitesimals in the ring of dual numbers are not invertible. So, it's the end of the party. (As you say though, some aspects of the party remain with automatic differentiation). In some sense, the dual numbers form a first order approximation to actual infinitesimals: The square of an infinitesimal is of an order of magnitude smaller than the infinitesimal you started with, but in the ring of dual numbers, the square of an 'infinitesimal' is precisely $0$. So, in a nonstandard model of the reals you have whole layers of infinitesimals. In the dual numbers there is only one layer, nothing in it is invertible, and they all square to $0$.
The book Models for smooth infinitesimal analysis explores many different models for analysis with infinitesimals. None of them is particularly simple.