Changing the Clifford square sign changes the low-rank algebra

For one generator $e$, the tensor algebra is $\mathbb{R}[t]$ and the Clifford relation is $t^2=Q(e)$. Division by either monic quadratic leaves a unique remainder $a+bt$. The positive evaluation map sends this remainder to $(a+b,a-b)$, which determines $(a,b)$, so $\mathbb{R}[t]\mathrm{/}(t^2-1)\cong \mathbb{R}\oplus \mathbb{R}$. The negative evaluation map sends it to $a+bi$, which determines $(a,b)$, so $\mathbb{R}[t]\mathrm{/}(t^2+1)\cong \mathbb{C}$.

The element $(1,0)$ is a nontrivial idempotent in $\mathbb{R}\oplus \mathbb{R}$. If $z=a+bi\in \mathbb{C}$ satisfies $z^2=z$, comparing real and imaginary parts gives $2ab=b$ and $a^2-b^2=a$. If $b=0$, then $a=0$ or $a=1$. If $b\mathrm{\ne }0$, then $a=1\mathrm{/}2$ and $b^2=-1\mathrm{/}4$, impossible because a real square is nonnegative. Thus $\mathbb{C}$ has no idempotents other than $0,1$. An algebra isomorphism preserves idempotents and $0,1$, so these algebras are not isomorphic.