Let $R = \{ b_0, b_1, \cdots, b_n \}$ be such a ring where $b_0 = 0$ and $b_1 = 1$.

If $n = 1$, we are done.

Suppose otherwise. All we need to show that each $b_i$ has a multiplicative inverse.

Let $i = 2, \cdots, n$ be given.

Set $f_i: b_j \mapsto b_i b_j$.

Suppose that for some $j, k = 1, \cdots, n$, $b_ib_j = b_ib_k$. This implies $b_i(b_j - b_k) = 0$, so $b_j - b_k = 0$.

Therefore, $f_i$ maps $\{ b_1, \cdots, b_n \}$ into itself.

This implies that $b_i b_j = b_1 = 1$ for some $j$. We were able to find the multiplicative inverse for any arbitrary non-zero element. Therefore, $R$ is indeed a field.