Considering a cylinder of radius $R$, the condition for rolling without slipping is met when $v=\omega R$. What we'll do is basically find how $\omega$ and $v$ vary in time and equate them:
To find $\omega$ we'll use the rotational analog of Newton's $2^{\text{nd}}$ law:
$$\begin{aligned}
\tau=||\mathbf r\wedge\boldsymbol f||=||(0,-R)\wedge(-f,0)||=fR=\mathcal I\alpha=\mathcal I\frac{\mathrm d\omega}{\mathrm dt}&\implies\int_0^\omega\mathrm d\omega'=\frac{fR}{\mathcal I}\int_0^t\mathrm dt'\\
&\implies \omega(t)=\frac{fR}{\mathcal I}t=\frac{2\mu g}{R}t,
\end{aligned}$$
where in the last step we used the fact that $\mathcal I_{\text{cylinder}}=\dfrac{1}{2}mR^2$ and $f=\mu N=\mu mg$.
As for $v$ we'll use Newton's $2^{\text{nd}}$ law
$$m\frac{\mathrm dv}{\mathrm dt}=-f=-\mu mg\implies\int_{v_0}^v\mathrm dv'=-\mu g\int_0^t\mathrm dt'\implies v(t)=v_0-\mu gt$$
Imposing $v(t_{\mathrm r})=\omega(t_{\mathrm r})R$, we get that
$$v_0-\mu gt_{\mathrm r}=2\mu gt_{\mathrm r}\implies t_{\mathrm r}=\frac{v_0}{3\mu g}$$
Finally to find the distance travelled by the cylinder once it starts rolling without slipping, we shall integrate $v$ once again an evaluate it at $t_{\mathrm r}$:
$$d=\int_0^{t_{\mathrm r}}v\mathrm dt=\left[v_0t-\frac{1}{2}\mu gt^2\right|_0^{t_{\mathrm r}}=v_0\left(\frac{v_0}{3\mu g}\right)-\frac{1}{2}\mu g\left(\frac{v_0}{3\mu g}\right)^2=\left(\frac{1}{3}-\frac{1}{18}\right)\frac{v_0^2}{\mu g}=\frac{5v_0^2}{18\mu g}$$
It might be doable by energies like you suggested, but it'd be a bit more tedious alternative to get to the same conclusion since what you're asking can't be simpler than imposing $v=\omega R$.
