Here are a couple of supplements to Omar's answer.

The first thing to note is that the module $\mathcal{E}_I$ is equal to the module $\mathcal{E}I=\{ei\ |\ e\in \mathcal{E},\ i\in I\}$. The inclusion $\mathcal{E}I\subset \mathcal{E}_I$ follows easily from the linearity of the inner product and from the fact that $I$ is an ideal. For the reverse inclusion, let $i_\lambda$ be an approximate unit for $I$, and then show (by expressing the norm in terms of the inner product) that for each $e\in \mathcal{E}_I$ one has $\| e-ei_\lambda\|\to 0$ as $\lambda\to\infty$. This gives $\mathcal{E}_I\subset \overline{\mathcal{E}I}$, and the latter is equal to $\mathcal{E}I$ by the Cohen factorisation theorem. For full details see Lemma 3.23 in:

*Raeburn, Iain; Williams, Dana P.*, Morita equivalence and continuous-trace $C^*$-algebras, Mathematical Surveys and Monographs. 60. Providence, RI: American Mathematical Society (AMS). xiv, 327 p. (1998). ZBL0922.46050.

Now, turning to the inner product on $\mathcal{E}/\mathcal{E}_I$: here are three arguments showing that the norm defined by the inner product is equal to the quotient norm.

(1) A short direct computation is given in Lemma 3.1 in:

*Zettl, Heinrich H.*, **Ideals in Hilbert modules and invariants under strong Morita equivalence of C*-algebras**, Arch. Math. 39, 69-77 (1982). ZBL0498.46034.

(2) An argument based on the uniqueness of the $C^*$-norm on the linking algebra is given in Proposition 3.25 in the book of Raeburn-Williams cited above. (The authors attribute the argument to Siegfried Echterhoff.)

(3) Here is a third proof, using ideas around operator modules and the Haagerup tensor product. All of the necessary background can be found in

*Blecher, David P.; Le Merdy, Christian*, Operator algebras and their modules -- an operator space approach, London Mathematical Society Monographs. New Series 30; Oxford Science Publications. Oxford: Oxford University Press (ISBN 0-19-852659-8/hbk). x, 387~p. (2004). ZBL1061.47002.

Consider $I\to A\to A/I$ as an exact sequence of operator $A$-bimodules. Taking the Haagerup tensor product with $\mathcal E$ (viewed as a right operator module over $A$) gives
$$
\mathcal E\otimes_A I \to \mathcal E\otimes_A A \to \mathcal E\otimes_A (A/I).
$$
By the exactness property of the Haagerup tensor product over a $C^*$-algebra (a theorem of Anantharaman-Delaroche and Pop), the first map in the display is a (completely) isometric embedding, and the second map induces a (completely) isometric isomorphism
$$
(\mathcal E\otimes_A A)/ (\mathcal E\otimes_A I) \cong \mathcal E\otimes_A (A/I)\qquad (*)
$$
where the left-hand side carries its canonical quotient operator space structure (and in particular, the usual Banach-space quotient norm).

The module action gives a (completely) isometric isomorphism $\mathcal E\otimes_A A \to \mathcal E$, which restricts to an isomorphism $\mathcal E\otimes_A I \to \mathcal E I=\mathcal E_I$. Making these identifications, the isomorphism $(*)$ is given by the formula
$$
\mathcal E/\mathcal E_I\to \mathcal E\otimes_A (A/I),\qquad (ea+\mathcal E_I)\mapsto e\otimes(a+I).\qquad (**)
$$
Now, $A/I$ is a (right) Hilbert $C^*$-module over itself, and the quotient map $A\to A/I$ gives a (left) action of $A$ on this $C^*$-module by adjointable operators. We can thus form the Hilbert $C^*$-module tensor product $\mathcal E\otimes^{C^*}_A (A/I)$, which will be a Hilbert $C^*$-module over $A/I$. A theorem of Blecher asserts that the identity map on the algebraic tensor products extends to a (completely) isometric isomorphism between $\mathcal E\otimes^{C^*}_A (A/I)$ and the Haagerup tensor product $\mathcal E\otimes_A (A/I)$. A straightforward computation shows that the map $(**)$ is isometric with respect to the inner product $[\cdot,\cdot]$ on $\mathcal E/\mathcal E_I$, and the canonical inner product on $\mathcal E\otimes^{C^*}_A (A/I)$. Since $(**)$ is also isometric for the quotient norm, the latter must coincide with the norm induced by $[\cdot,\cdot]$.