I am going to flesh out some of the comments to this post into an answer (with thanks to [Glen_b](https://stats.stackexchange.com/users/805/glen-b) and [AlexR](https://stats.stackexchange.com/users/78492/alexr) for their excellent comments).  The first thing to note is that the response variable in a logistic regression is binary, with allowable values $y_i = 0,1$, so any distribution that does not accord with that support is incorrect.  The distribution for the response variable in a logistic regression is Bernoulli, with a logistic mean.

> **Response distribution:** The response distribution for a logistic regression is:
> 
> $$Y_i | \mathbf{x} \sim \text{Bern}(\pi(x_i)) \quad \quad \quad \pi(x_i) = \frac{\exp(\beta_0+\beta_1 x_i)}{1 + \exp(\beta_0+\beta_1 x_i)}.$$
>
> The function $\pi$ is a [logistic function](https://en.wikipedia.org/wiki/Logistic_function) of an affine transformation of the argument value $x_i$.

The logistic regression model can be represented using various (equivalent) model forms that yield the above response distribution.  One of these model formulations uses a pseudo-error term with a standard [logistic distribution](https://en.wikipedia.org/wiki/Logistic_distribution).  Taking $\varepsilon_i \sim \text{Logistic}(0, 1)$ we have:

$$\begin{equation} \begin{aligned}
\mathbb{P}(\beta_0+\beta_1 x_i + \varepsilon_i > 0) = 1- F_{\varepsilon_i}(0) 
&= 1 - \frac{1}{1 + \exp(\beta_0+\beta_1 x_i)} \\[6pt]
&= \frac{\exp(\beta_0+\beta_1 x_i)}{1 + \exp(\beta_0+\beta_1 x_i)} \\[8pt]
&= \pi(x_i). \\[6pt]
\end{aligned} \end{equation}$$

Since this is the probability of a positive response outcome in the above distribution, we can formulate the logistic regression model in a way that is similar to a standard linear regression model, but with a pseudo-error term that is only used to classify the response into two categories:

$$Y_i = \mathbb{I}(\beta_0+\beta_1 x_i + \varepsilon_i > 0) \quad \quad \quad \varepsilon_1, ..., \varepsilon_n | \mathbf{x} \sim \text{IID Logistic}(0, 1).$$