In order to prove that any class $\mathrm{No}$ satisfying this definition is well-ordered by $\in$ it is crucial to use the axiom of foundation. Axiom of foundation tells us that $\in$ is a well-founded relation. Thus we only need to prove that any two elements of $\mathrm{No}$ are $\in$-comparable. This could be achieved by a proof by a contradiction, where we consider the $\in$-pointwise-minimal $\in$-incomparable pair $a,b\in\mathrm{No}$. This is essentially the same as the arguments that use ranks of sets suggested in the comments (one needs axiom of foundation to prove that every set has a rank), but formulated in the terms of more elementary notions.

If one drops the axiom of foundation, then we could have models, where there are classes satisfying this definition, although not being well-ordered by $\in$-relation. For example, it is possible to have a model with two chains of sets $a_i=\{a_j\mid j>i\}$ and $b_i=\{b_j\mid j>i\}$ such that all $a_i$ and $b_i$ are pairwise distinct sets. Clearly, $\{a_i\mid i\in\omega\}\cup \{b_i\mid i\in\omega\}$ satisfies your definition, but is neither linearly ordered nor well-founded.

In order to prove that any class $\mathrm{No}$ satisfying this definition is well-ordered by $\in$ it is crucial to use the axiom of foundation. Axiom of foundation tells us that $\in$ is a well-founded relation. Thus we only need to prove that any two elements of $\mathrm{No}$ are $\in$-comparable. This could be achieved by a proof by a contradiction, where we consider a $\in$-pointwise-minimal $\in$-incomparable pair $a,b\in\mathrm{No}$ to get to a contradiction. This is essentially the same as the arguments that use ranks of sets suggested in the comments (one needs axiom of foundation to prove that every set has a rank), but formulated in the terms of more elementary notions.

If one drops the axiom of foundation, then we could have models, where there are classes satisfying this definition, although not being well-ordered by $\in$-relation. For example, it is possible to have a model with two chains of sets $a_i=\{a_j\mid j>i\}$ and $b_i=\{b_j\mid j>i\}$ such that all $a_i$ and $b_i$ are pairwise distinct sets. Clearly, $\{a_i\mid i\in\omega\}\cup \{b_i\mid i\in\omega\}$ satisfies your definition, but is neither linearly ordered nor well-founded.

To prove that any $\mathrm{No}$ satisfying this definition is well-ordered by $\in$ it is crucial to use the axiom of foundation. Axiom of foundation tells us that $\in$ is a well-founded relation. Thus we only need to prove that any two elements of $\mathrm{No}$ are $\in$-comparable which could be achieved by a proof by contradiction, where we consider the $\in$ pointwise minimal incomparable pair $a,b\in\mathrm{No}$. This is essentially the same as the arguments that use ranks of sets suggested in the comments (one needs axiom of foundation to prove that every set has a rank), but formulated in terms of more elementary notions.

If one drops axiom of foundation, then we could have models, where there are classes satisfying this definition but that aren't well-founded. For example, it is possible to have a model with two chains of sets $a_i=\{a_j\mid j>i\}$ and $b_i=\{b_j\mid j>i\}$ such that all $a_i$ and $b_i$ are pairwise distinct sets. Clearly, $\{a_i\mid i\in\omega\}\cup \{b_i\mid i\in\omega\}$ satisfies your definition but is neither linearly ordered nor well-founded.

In order to prove that any class $\mathrm{No}$ satisfying this definition is well-ordered by $\in$ it is crucial to use the axiom of foundation. Axiom of foundation tells us that $\in$ is a well-founded relation. Thus we only need to prove that any two elements of $\mathrm{No}$ are $\in$-comparable. This could be achieved by a proof by a contradiction, where we consider the $\in$-pointwise-minimal $\in$-incomparable pair $a,b\in\mathrm{No}$. This is essentially the same as the arguments that use ranks of sets suggested in the comments (one needs axiom of foundation to prove that every set has a rank), but formulated in the terms of more elementary notions.

If one drops the axiom of foundation, then we could have models, where there are classes satisfying this definition, although not being well-ordered by $\in$-relation. For example, it is possible to have a model with two chains of sets $a_i=\{a_j\mid j>i\}$ and $b_i=\{b_j\mid j>i\}$ such that all $a_i$ and $b_i$ are pairwise distinct sets. Clearly, $\{a_i\mid i\in\omega\}\cup \{b_i\mid i\in\omega\}$ satisfies your definition, but is neither linearly ordered nor well-founded.