Noting that $f(x)g(y-x) = 0$ whenever $x < -1$ or $x > 1,$
\begin{align}
f \star g(y) &= \int_{-\infty}^\infty f(x)g(y-x)\, dx \\
&= \int_{-1}^1 f(x)g(y-x)\, dx \\
&= \int_{-1}^1 g(y-x)\, dx \\
&= -\int_{y+1}^{y-1} g(t)\, dt && t = y - x \\
&= \int_{y-1}^{y+1} g(t)\, dt.
\end{align}
But also $g(x)f(y-x) = 0$ whenever $y-x < -1$ or $y-x > 1$, that is to say,
whenever $x > y+1$ or $x < y-1,$ so
\begin{align}
g \star f(y) &= \int_{-\infty}^\infty g(x)f(y-x)\, dx \\
&= \int_{y-1}^{y+1} g(x)f(y-x)\, dx \\
&= \int_{y-1}^{y+1} g(x)\, dx .
\end{align}
The final value of the integral either way is not as simple as
$-2y^2 + \frac43.$
If $y \leq -2$ or $y \geq 2$ the integral is zero.
If $y = 0$ then
\begin{align}
g \star f(y) &= \int_{-\infty}^\infty g(x)f(-x)\, dx \\
&= \int_{-1}^{1} g(x)\, dx \\
&= \int_{-1}^{1} (1 - x^2)\, dx \\
&= 2 - \left. \frac{x^3}{3}\right\rvert_{x=-1}^{x=1} \\
&= \frac 43.
\end{align}
But if $0 < y < 2$ then $f(y - x) = 0$ whenever $x < y - 1,$
and $g(x) = 0$ whenever $x > 1,$ so
\begin{align}
g \star f(y) &= \int_{-\infty}^\infty g(x)f(y-x)\, dx \\
g \star f(y) &= \int_{y-1}^{1} g(x) \, dx \\
g \star f(y) &= \int_{y-1}^{1} (1 - x^2)\, dx \\
&= (2 - y) - \left. \frac{x^3}{3}\right\rvert_{x=y-1}^{x=1} \\
&= \frac{y^3}{3} - y^2 + \frac 43.
\end{align}
And if $-2 < y < 0$ then $f(y - x) = 0$ whenever $x > y + 1,$
and $g(x) = 0$ whenever $x < -1,$ and the result is
\begin{align}
g \star f(y) &= -\frac{y^3}{3} - y^2 + \frac 43.
\end{align}
So $g\star f(y)$ depends very much on $y$.
Moreover, your calculation of $f\star g(y)$ is incorrect.
You need to account for the fact that $g(x)=1 - x^2$ on all of $[y-1,y+1]$ only when $y = 0.$ For $y > 0,$ $g(x)=1 - x^2$ on $[y-1,1]$ and $g(x) = 0$ otherwise.
Indeed, you should end up integrating the exact same integrals as for $g\star f.$
