What proportion of your 225,000 training samples are cases of fraud? I suspect very few. This will cause problems unless you take care in how you build a classifier from the logistic regression.
Given the issue you've described, I assume you are making your classifications based on a cut-off of a probability of $0.5$ from the logistic regression. You either need to choose a more appropriate cut-off, or weight the fraud samples (a weight of $19$ would be appropriate given you have $19$ times more not fraud cases in your data set), or discard a lot of the not fraud cases so you have a balanced data set.
As for your second question, to properly assess the out of sample performance I would calculate the average/min/max of each column only using your training set. As an alternative to scaling you might consider binning the relevant variables.
The two metrics measure slightly different things.
Naive Bayes tries to learn how to approximate $P(\textrm{class } | \textrm{ data})$ from your training data. To classify new data, the algorithm takes each example and computes $P(\textrm{class}=c\ | \textrm{ data})$ for each possible class $c$. We label each test example with the $c$ that maximizes that probability.
The accuracy metric only concerns itself with final classification output. For each example in the test set, you get a 1 if the largest $c$ is the correct class, and a zero otherwise. Average these values together and you'll get an accuracy measurement. In other words, accuracy is 0/1 loss, scaled into a percentage. Log-loss, however, considers the probability $P(\textrm{class}=c\ | \textrm{ data})$ directly. It's essentially the sum of the negative log-likelihoods of the true classes given your model. Kaggle defines it slightly differently, but in either case, the actual probabilities matter, not just which class's probability is the largest.
Here's an example. Suppose your first model is stupid but noncommittal. For 9 examples, it estimates $P(\textrm{true class})=0.45$ and $P(\textrm{other}) =0.55$, while it does the opposite for one example. This yields an accuracy of 10% and a log-loss of $\frac{1}{10}\big(9\log(0.45) + 1\log(0.55)\big) = 0.78$.
The second model is better but also more decisive, even when it's wrong. Suppose it calculates $P(\textrm{true class})=0.9$ for two examples, and $P(\textrm{true class})=0.1$ for the remaining eight. This classifier has an accuracy of 20%, but a log-loss of 1.8, which mimics your situation exactly.
Best Answer
I assume you are using Gaussian, not Multinomial/Binomial Naive Bayes?
For Gaussian Naive Bayes, the estimator learns the mean and standard deviation of each feature (per class). At prediction time the probability of a value being in a class is a function of the distance from the center of the distribution. The function used is Probability Density Function (PDF), of a Normal/Gaussian distribution. And the Normal PDF is just a Standard Normal distribution (0 mean, unit variance) that is scaled by variance and shifted by mean. So a value which is at mean+(0.5*std) has the same probability.
With standardization the mean and stddev changes, but probabilities stay exactly the same, and thus classification results. In essence Gaussian Naive Bayes performs standardization internally.