(I) When $a=-1$, we have $f(x)=|2x+1|+|2x-1|$. The inequality $f(x) \leqslant 2$ implies $|x+\frac{1}{2}|+|x-\frac{1}{2}| \leqslant 1$.

This inequality represents the sum of distances on the number line from point $x$ to two points $(-\frac{1}{2})$ and $(\frac{1}{2})$ being less than or equal to $1$. Therefore, $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$.

Hence, the solution set for the original inequality is $\boxed{\left[-\frac{1}{2}, \frac{1}{2}\right]}$.

(II) Since the solution set for $f(x) \leqslant |2x+1|$ contains $\left[\frac{1}{2}, 1\right]$, the inequality $f(x) \leqslant |2x+1|$ must always hold true when $x \in \left[\frac{1}{2}, 1\right]$.

Thus, for all $x \in \left[\frac{1}{2}, 1\right]$, we have $|2x-a| + 2x - 1 \leqslant 2x + 1$,

which simplifies to $|2x-a| \leqslant 2$. This leads to $-2 \leqslant 2x - a \leqslant 2$, or $2x - 2 \leqslant a \leqslant 2x + 2$.

For this inequality to hold true for all $x \in \left[\frac{1}{2}, 1\right]$, we must have $\max(2x-2) \leqslant a \leqslant \min(2x+2)$. Thus, $0 \leqslant a \leqslant 3$.

Hence, the possible values for $a$ are $\boxed{[0, 3]}$.