The formal "defintion" of stochastic differential is what you wrote at the beginning: $dX_t=...$ is a short form of $X_t=X_0+...\,.$ In particular $ds_t$ has no formal definition other than the equation wrote: $s_t=s_0+ct+Z_t\,.$ The rest follows from $\mathbb E_t[Z_t]=Z_t$ and $\mathbb E_t[Z_{t+u}]=Z_t$ and $$\textstyle\mathbb E_t[ct]=-\frac{\sigma^2t}{2}\,,\quad \mathbb E_t[c(t+u)]=-\frac{\sigma^2(t+u)}{2}$$ assuming that $c\sim N(-\frac{\sigma^2}{2},\sigma^2)$ is independent of the Brownian motion $Z\,.$

These formulas could be used to calculate the derivative $$\tag{1} \frac{d}{dT}\Bigg|_{T=t}\mathbb E_\color{red}{t}\big[s_T\big]=-\frac{\sigma^2}{2}\,. $$ Imho this is no reason to conclude that we can formally define the conditional expectation $$\tag{2} \mathbb E_\color{red}{t}\big[ds_t\big] $$ of $ds_t\,.$ What should the random variable $ds_t$ be? If it is ${\cal F}_t$-measurable then $\mathbb E_\color{red}{t}\big[ds_t\big]=ds_t$ - in contrast to the other answer.

The minimal compromise I accept is that $$ \mathbb E_\color{red}{t}\big[ds_t\big]=-\frac{\sigma^2}{2} $$ is an "abbreviation" of (1), albeit a very confusing one.

The formal "defintion" of stochastic differential is what you wrote at the beginning: $dX_t=...$ is a short form of $X_t=X_0+...\,.$ In particular $ds_t$ has no formal definition other than the equation wrote: $s_t=s_0+ct+Z_t\,.$ The rest follows from $\mathbb E_t[Z_t]=Z_t$ and $\mathbb E_t[Z_{t+u}]=Z_t$ and $$\textstyle\mathbb E_t[ct]=-\frac{\sigma^2t}{2}\,,\quad \mathbb E_t[c(t+u)]=-\frac{\sigma^2(t+u)}{2}$$ assuming that $c\sim N(-\frac{\sigma^2}{2},\sigma^2)$ is independent of the Brownian motion $Z\,.$

These formulas could be used to calculate the derivative $$\tag{1} \frac{d}{dT}\Bigg|_{T=t}\mathbb E_\color{red}{t}\big[s_T\big]=-\frac{\sigma^2}{2}\,. $$ Imho this is no reason to conclude that we can formally define the conditional expectation $$\tag{2} \mathbb E_\color{red}{t}\big[ds_t\big] $$ of $ds_t\,.$ What should the random variable $ds_t$ be? If it is ${\cal F}_t$-measurable then $\mathbb E_\color{red}{t}\big[ds_t\big]=ds_t$ - in contrast to the other answer.

The minimal compromise I accept is that $$ \mathbb E_\color{red}{t}\big[ds_t\big]=-\frac{\sigma^2}{2}\,dt $$ is an "abbreviation" of (1), albeit a very confusing one.

The formal "defintion" of stochastic differential is what you wrote at the beginning: $dX_t=...$ is a short form of $X_t=X_0+...\,.$ In particular $ds_t$ has no formal definition other than the equation wrote: $s_t=s_0+ct+Z_t\,.$ The rest follows from $\mathbb E_t[Z_t]=Z_t$ and $\mathbb E_t[Z_{t+u}]=Z_t$ and $$\textstyle\mathbb E_t[ct]=-\frac{\sigma^2t}{2}\,,\quad \mathbb E_t[c(t+u)]=-\frac{\sigma^2(t+u)}{2}$$ assuming that $c\sim N(-\frac{\sigma^2}{2},\sigma^2)$ is independent of the Brownian motion $Z\,.$

The formal "defintion" of stochastic differential is what you wrote at the beginning: $dX_t=...$ is a short form of $X_t=X_0+...\,.$ In particular $ds_t$ has no formal definition other than the equation wrote: $s_t=s_0+ct+Z_t\,.$ The rest follows from $\mathbb E_t[Z_t]=Z_t$ and $\mathbb E_t[Z_{t+u}]=Z_t$ and $$\textstyle\mathbb E_t[ct]=-\frac{\sigma^2t}{2}\,,\quad \mathbb E_t[c(t+u)]=-\frac{\sigma^2(t+u)}{2}$$ assuming that $c\sim N(-\frac{\sigma^2}{2},\sigma^2)$ is independent of the Brownian motion $Z\,.$

These formulas could be used to calculate the derivative $$\tag{1} \frac{d}{dT}\Bigg|_{T=t}\mathbb E_\color{red}{t}\big[s_T\big]=-\frac{\sigma^2}{2}\,. $$ Imho this is no reason to conclude that we can formally define the conditional expectation $$\tag{2} \mathbb E_\color{red}{t}\big[ds_t\big] $$ of $ds_t\,.$ What should the random variable $ds_t$ be? If it is ${\cal F}_t$-measurable then $\mathbb E_\color{red}{t}\big[ds_t\big]=ds_t$ - in contrast to the other answer.

The minimal compromise I accept is that $$ \mathbb E_\color{red}{t}\big[ds_t\big]=-\frac{\sigma^2}{2} $$ is an "abbreviation" of (1), albeit a very confusing one.