Let $\mu, \sigma$ be constants and $(B_t)_t$ a brownian motion and define the process $S_t=S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma B_t}$. Then define $U_t=\frac{1}{S_t}$. I want to find a process $(A_t)$ s.t. $U_t=\mathcal{E}(A)_t$. Where $\mathcal{E}(A)_t$ denotes the stochastic exponential.
My idea was the following:
I know that $S_t$ solves the SDE $dS_t=\mu S_t dt+\sigma S_tdB_t$. And I also know that for a general process $(X_t)$ we have $\mathcal{E}(X)_t=\exp\left(X_t-\frac{1}{2}\langle X\rangle_t\right)$ which solves the SDE $\mathcal{E}(X)_t=\mathcal{E}(X)_0+\int_0^t \mathcal{E}(X)_s dX_s~~(1)$.
Now let me define $f(x,s)=\frac{1}{x}$ then $\partial_xf(x,s)=-\frac{1}{x^2}$, $\partial_{xx}f(x,s)=\frac{2}{x^3}$, $\partial_sf(x,s)=0$. So by Itos formula we get $$\begin{align}U_t&=f(S_0,0)+\int_0^t \partial_xf(S_s,s)dS_s+\frac{1}{2}\int_0^t\partial_{xx}f(S_s,s)d\langle S\rangle_s\\&=\frac{1}{S_0}-\mu \int_0^t f(S_s,s)ds+\sigma \int_0^tf(S_s,s)dB_s+\sigma^2 \int_0^t f(S_s,s)ds\\&=f(S_0,0)-(\mu-\sigma^2)\int_0^tf(S_s,s)ds-\sigma \int_0^t f(S_s,s)dB_s\\\end{align}$$ Now my idea was to define $dA_t=-(\mu-\sigma^2)dt-\sigma dB_t$ because then we can rewrite $f(S_t,t)=f(S_0,0)+\int_0^tf(S_s,s)dA_s$ and hence we know by the "characterisation" of stochastic exponential $(1)$ that $U_t=f(S_t,t)=\mathcal{E}(A)_t$.
But does this idea work or am I wrong?
Too long for a comment:
You are overthinking it. First let me remark that the solution of the SDE $dS_t=\mu S_t\,dt+\sigma S_t\,dB_t$ is $$ S_t=S_0\,e^{\mu t-\frac{\sigma^2}{2}t+\sigma B_t} $$ which differs from your $S_t\,.$ Then, $$ \textstyle U_t=\frac1{S_t}=\frac1{S_0}e^{-\mu t+\frac{\sigma^2}{2}t-\sigma B_t}\,. $$ With your definition ${\cal E}(X)_t=e^{X_t-\langle X\rangle_t/2}$ we can obviously choose $$ X_t=-\log S_0-\mu t+\sigma^2t-\sigma B_t\, $$ to have $U_t={\cal E}(X)_t\,.$