Where have I gone wrong?
You've gone wrong here:
the resistance of some component is equal to the total voltage
difference through that component divided by the total current that
flows through it.
It's not some component that Ohm's law applies to, it's specifically (ideal) resistors.
To model a physical cell, one starts by putting an ideal resistor in series with an ideal voltage source. You'll note that the voltage across the resistor is given by Ohm's law but the voltage across the voltage source is what it is regardless of the current through, i.e., the ideal voltage source does not obey Ohm's law.
Thus, the voltage across the series combination of the voltage source and resistor will not, and should not be expected to, obey Ohm's law.
Seeing the updates to your original post, I think I better see what you're asking about.
As I wrote above, Ohm's law $V = I R$ where $R$ is a constant applies to ideal resistors.
However, for non-ohmic components, one can define a static (or DC) resistance $R_{DC}$ as well as a dynamic (or small-signal) resistance $r$.
The static resistance is simply the ratio of the DC voltage across and current through:
$$R_{DC} \equiv \frac{V_{DC}}{I_{DC}}$$
So, in the battery example of your question, the static resistance is given by
$$R_{DC} = \frac{E + I_{DC}r}{I_{DC}} = \frac{E}{I_{DC}} + r$$
(Note: this is a somewhat unusual application of the static resistance concept since the battery is a typically a source rather than a load).
The The dynamic resistance is the ratio of the change in voltage to the change in current from their DC values:
$$r \equiv \frac{dV}{dI} \approx \frac{\Delta V}{\Delta I}$$
In your battery example, the dynamic resistance is just the internal resistance $r$
Best Answer
A real am-meter (like the one on your desk) has a very small resistance $R$ (may be $0.001\ \Omega$). The current ($I$) trough your am-meter and the potential difference ($V$) across it are related by Ohm's law: $$V=RI$$ That means, for any current $I$ you have a small potential difference $V$.
On the other hand, an ideal am-meter (which, as the name says, is an idealization, and does not really exist) has a zero resistance $R=0$. Then Ohm's law from above becomes $$V=0\cdot I$$ or $$V=0.$$
That means, the current $I$ through the am-meter can be anything, and the potential difference $V$ across the am-meter is zero. So there is no contradiction between $V=0$ and $I\neq 0$.