Do we need the axiom of replacement (ZFC) to define a product of structures (Universal Algebra)?
Here $\mathrm{V}$ is the class of all sets.
From my perspective here is how the product of structures is defined.
Definitions:
Let $ X $ be a set. An arity-relation in $ X $ is a 2-tuple $ \left(\vphantom{n}\smash{\breve{n}},\asymp\right) $, where $ \vphantom{n}\smash{\breve{n}}\in \mathbb{N} $ and $ {\asymp}\subseteq X^{\vphantom{n}\smash{\breve{n}}} $.
Let $ X $ be a set. An arity-operation on $ X $ is a 2-tuple $ \left(\vphantom{n}\smash{\dot{n}},\odot\right) $, where $ \vphantom{n}\smash{\dot{n}}\in \mathbb{N} $ and $ {\odot}:X^{\vphantom{n}\smash{\dot{n}}}\to X $.
A structure is a 3-tuple $ \left(X,(\vphantom{n}\smash{\breve{n}},\asymp),(\vphantom{n}\smash{\dot{n}},\odot)\right) $, where $ X $ is a set and $ \{(\vphantom{n}\smash{\breve{n}}_{(\cdot)},\asymp_{(\cdot)})\} $ is a family of arity-relations in $ X $ and $ \{(\vphantom{n}\smash{\dot{n}}_{(\cdot)},\odot_{(\cdot)})\} $ is a family of arity-operations on $ X $.
Let $ \{(X_\lambda,(m,\asymp^{\lambda}),(n,\odot^{\lambda}))\}_{\lambda\in\Lambda} $ be a family of structures. Fix nonempty $ I\subseteq \Lambda $. The product of $ (X_i,(m,\asymp^i),(n,\odot^i)) $ for $ i\in I $, denoted $ \prod_{i\in I}(X_i,(m,\asymp^i),(n,\odot^i)) $, is the 3-tuple $ \left(Y,(m,\vphantom{\asymp}\smash{\bar{\asymp}}),(n,\vphantom{\odot}\smash{\bar{\odot}})\right) $, where $ Y=\prod_{i\in I}X_i $ and
$ \vphantom{\asymp}\smash{\bar{\asymp}}_{(\cdot)}:\operatorname{dom}\left(m\right)\to \mathrm{V}:\alpha\mapsto \{y^{(\cdot)}\in Y^{m_\alpha}:\left(\forall i\in I\right)[\pi_i\circ y\in ({\asymp}^i)_\alpha]\} $ and
$ \vphantom{\odot}\smash{\bar{\odot}}_{(\cdot)}:\operatorname{dom}\left(n\right)\to\mathrm{V}:\alpha\mapsto \{(y^{(\cdot)},z)\in Y^{n_\alpha}\times Y:\left(\forall i\in I\right)[(\odot^i)_\alpha(\pi_i\circ y)=z_i]\} $.
Question: Can the codomain of $\bar{\asymp}_{(\cdot)}$ or $\bar{\odot}_{(\cdot)}$, currently written as $\mathrm{V}$, be described without the axiom of replacement?
Context: I am interested in the axiom of replacement because of how little it is required in everyday mathematics (embedded in ZFC). Originally I was looking into Birkhoff's HSP Theorem in Universal Algebra where no relations beyond equality is in a structure. I simply extended the definition to include relations because that is what Model Theory studies. I provide this context because I received some downvotes on this question without explanation.
EDIT1: I want to add that I am not commited to the definition above. If you know of a different definition that can work in ZFC (minus replacement), then I will consider that an answer to the titular question. For example, the definition of an ordered pair $(a,b):=\{\{a\},\{a,b\}\}$ commonly used has the benefit of the separation axiom and power set axiom (among others) being enough to define products of sets. If one insists to define ordered pairs in an unknown fashion that so that $(a,b)=(x,y)\iff a=x\text{ and }b=y$, then the replacement axiom would be required to do much of the same things. For example, see $\text {dom}(R)$ and $\text {ran}(R)$ exists for any definition of order pair.
Every element of the image of $\bar{\asymp}_{(\cdot)}$ or $\bar{\odot}_{(\cdot)}$ is a subset of $Y^n$ for some $n\in\mathbb{N}$. That is, it is a set of functions $n\to Y$. Every such function for any $n$ can be considered as a relation (more specifically, a partial function) from $\mathbb{N}$ to $Y$. So, $\bar{\asymp}_{(\cdot)}$ can be considered to have codomain $\mathcal{P}(\mathcal{P}(\mathbb{N}\times Y))$, with no Replacement needed.