For an arbitrary semimartingale $Y_t$ show that $\mathcal{E}(\mathcal{L}(Y_t)) = Y_t/Y_0$.
I saw this post and tried to do the same for a non necessarily continuous process but can not really see the equality of the terms involving the jumps of the process. Hence I tried to use the other characterisation of the stochastic logarithmic and the exponential functions in terms of their differentials. Namely, that $W_t = \mathcal{E}(Y_t)$ is the process that satisfies $dW_t = W_t dY_t$ and if $W_t = \mathcal{L}(Y_t)$ then it satisfies $dW_t = dY_t/Y_t$.
Denoting $X_t = \mathcal{E}(\mathcal{L}(Y_t))$, using this differential characterisation it is easy to check that:
$$\frac{dX_t}{X_t} = \frac{dY_t}{Y_t}$$
or that $\mathcal{L}(Y_t) = \mathcal{L}(X_t)$ but possibly differing at their initial point. Can we conclude from this our claim? Can anybody add some comments on it?
The linked post defines neither the stochastic exponential, nor the stochastic logarithm of a semi-martingale, henceforth abbreviated by SM:
The stochastic (aka Doleans-Dade) exponential of a SM $X_t$ is the unique strong solution to the SDE $$ dY_t=Y_{t-}\,dX_t\,, \quad Y_0=1\, $$ and is denoted by $Y_t={\cal E}(X_t)\,.$
The stochastic logarithm of the SM $Y_t$ is the SM $X_t$ given by $$ dX_t=\frac{dY_t}{Y_{t-}}\,,\quad X_0=0\, $$ and is denoted by $X_t={\cal L}(Y_t)\,.$
The claim $$\tag{1} \boxed{\phantom{\Big|}\quad {\cal E}({\cal L}(Y_t))=Y_t\quad} $$ is now trivial by plugging the definitions into each other.
In your notation \begin{align} X_t&={\cal E}(W_t)\,,\quad &dX_t&=X_{t-}\,dW_t\,,\;&X_0=1\,,\\[2mm] W_t&={\cal L}(Y_t)\,,\quad &dW_t&=\frac{dY_t}{Y_{t-}}\,,\;&W_0=0\,. \end{align} In particular $$ dY_t=Y_{t-}\,dW_t\,. $$ It follows that $X_t=Y_t$ holds as soon as we require $Y_0=1$ because that SDE has a unique strong solution. This implies again (1).