The function $\operatorname{W}_0(x)$ determines the solution of $$ye^y=x$$ with $y> -1$ when $-\frac{1}{e}<x<0$, whereas $\operatorname{W}_{-1}(x)$ determines the solution with $y< -1$
Is there any relation between $\operatorname{W}_{-1}(x)$ and $\operatorname{W}_0(x)$ that allows to calculate $\operatorname{W}_{-1}(x)$ , when $\operatorname{W}_0(x)$ is known ?
As far as I am aware the answer to your question seems to be no.
The best we can do is rewrite one of the known simplification formulae for the Lambert W function in a form suggestive of what you want. If $s > -1, s \neq 0$ and \begin{equation} a = \frac{\ln (1 + s)}{s(1 + s)^{1/s}}, \qquad (*) \end{equation} where we see we have $0 < a \leqslant 1/e$, one can write $$\text{W}_{-1} (-a) = (1 + s) \text{W}_0 (-a).$$
I say this equation is only suggestive of what you want since the $a$ you plug into the equation is not any old value for $a$ between zero and $1/e$ but depends on the value of $s$ given above by Eq. ($*$).