Quotient ring with $1\neq 0$?

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A quotient ring of a ring $R$ is (afaik) always formed by quotienting over an ideal of $R$.

But if I understand correctly, this causes the equivalence class containing all elements of the ideal to be both the additive and multiplicative identity.

Is there a way, perhaps using a different notion of "quotient ring" to form a quotient ring in which the additive and multiplicative identity are not the same equivalence class?

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But if I understand correctly, this causes the equivalence class containing all elements of the ideal to be both the additive and multiplicative identity.

No, it doesn't. In the quotient $R/I$, the additive identity is the equivalence class $I$ and the multiplicative identity is $1+I$. Those are different unless $I$ contains $1$ - and if $I$ contains $1$, it's all of $R$ and the quotient is the trivial ring $\{0\}$.