From the conditions given, we can infer that $a \neq 0$, as otherwise, the function $y=ax+2a+1$ would not be able to take on both positive and negative values (it would be a constant function).

Let us consider $y=f(x)=ax+2a+1$. For $y$ to take both positive and negative values between $x=-1$ and $x=1$, it's required that the product of the function values at these endpoints is negative. Thus $f(-1) \cdot f(1) 1$, both factors are positive again, giving a positive product.

Hence, the function takes both positive and negative values for $-\frac{1}{3} < a < 1$. But since $a$ cannot be equal to zero, we further refine the range to:
$$-1 < a < -\frac{1}{3}.$$

\boxed{-1 < a < -\frac{1}{3}}