Your reasoning is correct.
If the solar cell is modelled as a voltage source $E$ in series with an internal resistance $r$ and the cell is connected to a load resistance $R_L$, the series current is given by Ohm's law:
$$I = \frac{E}{r + R_L}$$
or
$$E = (r + R_L)\cdot I $$
The output voltage $V$ of the solar cell is the voltage across the load resistance which is, by Ohm's law, $V = R_L\cdot I$.
Thus
$$V = R_L\cdot I = E - r\cdot I $$
UPDATE :
John : Thanks for data. Graph is ok. I note your intercept is E=3.94V but your calculations use E=4.5V. This explains the discrepancy in your results. If you use 3.94V you get r ranging from 1.59 to 1.76, close to slope value of 1.68 Ohms.
ORIGINAL ANSWER :
Your line of best fit gives an average internal resistance r based on all measurements. If data points do not lie exactly on this line then the value of r calculated for individual data points (measured pairs of V and I) will not be exactly the same as the slope of the line of best fit.
If you have drawn the line correctly some points will be above the line and some below, with about as many each side, and with the above and below points distributed randomly.
However, it sounds as though there is a consistent trend in your data points : eg all 'below' points at low current and all 'above' points at high current. This suggests that internal resistance was not in fact constant, within the limitations of experimental error. You do not say how big an effect this is : if small, you may be able to ignore it.
EMF and r should be measured when the current drawn is very small, ideally 0. Possibly you have taken readings at a high current, or you have taken a long time to take them. This can have two effects : (i) depleting the battery, reducing EMF, and (ii) increasing r because the battery is warming up and this increases internal resistance.
Your observation that internal resistance increased as current decreased suggests to me that you may have started readings with a high current then worked down to low current.
You will need to decide for yourself what went wrong, perhaps after consulting your teacher again and explaining how you took the readings.
Best Answer
As you said,
then, remember that the current $I$ is:
$I=V/R$
so the current did not change in that branch. The total current will decrease though, because the current through the other branch did diminish (same equation, but $R $ increases).