Can we construct $\Bbb C$ without first identifying $\Bbb R$?

If you want to avoid $\Bbb R$ and just use general machinery, one way to do it is to use $\Bbb Q(i)$ or any finite extension of $\Bbb Q$ which has zero real embeddings. You can ensure this by taking the extension to be cyclotomic, for example. Then you know there is a norm on the vector space $\Bbb Q(i)$ given by

$$\lVert a+bi\rVert=|a|+|b|.$$

It's easily verified that it is archimedean--this is handy because it will give you a copy of $\Bbb R$ as a subset when you're finished making $\Bbb C$. Now, you can verify that addition, subtraction, multiplication, and inversion of non-zero elements is continuous so that you have a topological field.

Then by forming the metric completion and declaring it to be $\Bbb C$, you automatically have that this is a field because of continuity of the field operations. It is not otherwise obvious that your set of equivalence classes should form such a thing.

Any algebraically closed field of characteristic $0$ having the same cardinality as $\mathbb C$ is isomorphic (non-canonically) to $\mathbb C$. This allows one to construct a lot of fields which are abstractly isomorphic to $\mathbb C$ without ever looking at the real numbers.

For example, the algebraic closure of $\mathbb Q_p$ is isomorphic to $\mathbb C$. However, no isomorphism between them is continuous, so this is not very interesting if you're interested in $\mathbb C$ as a topological field.