A local ring $(R,\mathfrak{m})$ with $\operatorname{depth} R \ge \mu_R(\mathfrak{m})-1$ is often termed an abstract hypersurface (note this includes the case $\operatorname{depth} R=\mu(\mathfrak{m})$ i.e. when $R$ is regular). Sometimes this condition is defined instead to be that the completion is a hypersurface in the absolute sense, i.e., that $\hat{R} \cong S/(f)$ for some regular local ring $S$ and $f \in S$. That these are equivalent will be shown below. In some sense, this second definition is more natural since one defines an abstract complete intersection in the analogous manner.
We observe the following:
Suppose $R$ is an abstract hypersurface and suppose $R$ is the homomorphic image of a regular local ring. Then $R$ is a hypersurface in the absolute sense, i.e., $R \cong S/(f)$ for some regular local ring $S$ and $f \in S$.
Proof: We may write $R \cong S/I$ where $(S,\mathfrak{n})$ is a regular local ring and $I \subseteq \mathfrak{n}^2$. By assumption, $\operatorname{depth} S-\operatorname{depth} R=\mu_R(\mathfrak{m})-\operatorname{depth} R \le 1$, and thus, by Auslander-Buchsbaum, $\operatorname{pd}_S R \le 1$. If $\operatorname{pd}_S R=0$, then $R \cong S$ and we are done. Otherwise, $\operatorname{pd}_S R=1$. But we have an exact sequence $0 \to I \to S \to R \to 0$. As $\operatorname{pd}_S R=1$, this forces $I$ to be a free $S$-module, and thus $I$ is principal.
In particular, the above theorem shows that the completion of any abstract hypersurface is always a hypersurface in the absolute sense, since $\mu_R(\mathfrak{m})$ and $\operatorname{depth} R$ are preserved by completion, and thus in turns shows that abstract hypersurfaces enjoy many of the same homological properties enjoyed by hypersurfaces in the absolute sense; in particular they are always Gorenstein.
To answer your question directly, no; an example was given by Heitmann and Jorgensen of an abstract hypersurface which is not a hypersurface in the absolute sense. See the (excellently titled) paper ``Are complete intersections complete intersections?" https://arxiv.org/pdf/1109.4921.pdf.
Your argument is very useful. I follow your argument. Since $R/I$ has finite length, we know $$R/I\cong \mathrm{Hom}_R(\mathrm{Hom}_R(R/I,E_R(k)),E_R(k))=\mathrm{Hom}_R(N,E_R(k)).$$ You have showed that $\mathrm{pd}_RN<\infty$, so we know $\mathrm{id}_R(R/I)<\infty$.
Lemma 1. Let $(R,m,k)$ be a Noetherian local ring. If $M$ is a finitely generated $R$-module, then $\mathrm{id}_RM=\sup\{i\mid \mathrm{Ext}^i_R(k,M)\neq 0\}$.
By lemma 1, and by Nakayama we know:
Lemma 2. Let $(R,m,k)$ be a Noetherian local ring, $M$ is a finitely generated $R$-module, and $x\in m$ is a regular element on $M$. Then $\mathrm{id}_R(M)=\mathrm{id}_R(M/xM)$.
So we know $\mathrm{id}_R(R)=\mathrm{id}_R(R/I)<\infty$, since $I$ is generated by a regular sequence.
More generally, Foxby showed:
If $R$ is a Noetherian local ring, if there exists a finitely generated module with finite injective dimension and finite projective dimension, then $R$ is Gorenstein.
Best Answer
No, it is not true. It is possible to construct several examples using numerical semigroups. A numerical semigroup $S$ is simply a submonoid of $\mathbb{N}$ such that $\mathbb{N}\setminus S$ is finite. It is easy to see that every numerical semigroup has a unique minimal system of generators, which is finite. In this case I write $S=\langle s_1, \dots, s_\mu\rangle=\{\lambda_1 s_1 + \dots +\lambda_\mu s_\mu \mid \lambda_1, \dots, \lambda_\mu \in \mathbb N\}$.
Consider the ring $k[[S]]=k[[t^{s_1}, \dots, t^{s_\mu}]] \subseteq k[[t]]$, where $k$ is a field and $t$ is an indeterminate. This is a complete local Cohen-Macaulay ring of dimension one (it is indeed a domain). It is well known that in these rings the multiplicity equals the smallest non-zero element of $S$, while the embedding dimension is equal to the number of minimal generators, which is $\mu$ for us. Moreover, if $k[[S]]$ has minimal multiplicity, then the type is always $\mu-1$. Hence, every $S$ with at least three minimal generators and such that $k[[S]]$ has minimal multiplicity is a counterxample to your question.
For instance, for every $n \geq 3$ the ring $k[[t^n, t^{n+1}, \dots, t^{2n-1}]]$ has multiplicity and embedding dimension equal to $n$, whereas its type is $n-1 > 1$, and then it is not Gorenstein.