When fitting a model — say, finding weights (coefficients) in a linear regression — the loss function is used to determine the quality of a fit. Let's say that the loss function for a set of weights $\boldsymbol{w}$ depends on the absolute difference between the model prediction $\hat{x}$ and observation $\boldsymbol{x}$ in feature space, e.g.,
$$ loss(\boldsymbol{w}) = \sum_{i=1}^N (\hat{x_i}(\boldsymbol{w})-x_i)^2 $$
where index $i$ indicates the $i^{th}$ feature in the model. In this case, a feature with a very large scale (say, its values range from -2000 to 2000) would be far more influential in the fit than a feature with a very small scale (say, its values range from -0.01 to 0.01). In this case, feature scaling — e.g., z-standardizing the values of all of our features — should help us converge on the right fit.
However, let's say that we're looking at a classifier with a loss function that depends upon the feature vector $\boldsymbol{x}$ only through:
$$ \sigma(\boldsymbol{w}\cdot\boldsymbol{x}) \equiv \frac{1}{1+\exp(-\boldsymbol{w}\cdot \boldsymbol{x})} $$
where $w$ is a set of weights. This might occur, for example, when looking at the LogLoss for a binary classifier, in which for the $t^{th}$ observation and labels $y_t \in \{0,1\}$:
$$ loss(\boldsymbol{w}_{t}) = -y_t \log p_t - (1-y_t )\log (1-p_t) $$
and $p_t = \sigma(\boldsymbol{w}\cdot \boldsymbol{x})$. In this case, it appears that the weights can 'absorb' the scaling of individual features without an issue, preventing that scale from affecting the fit.
Question: Am I correct in thinking that in such cases, feature scaling is not necessary in order to achieve a good fit? Are there other features of the model that might suffer if we fail to scale our features?