In my answer I am using the convention that the complex logarithm $\log$ is a holomorphic function (the "principal branch" of $Log$) defined on
$$
{\mathbb C}\setminus (-\infty,0],
$$
where $(-\infty,0]\subset {\mathbb R}$. Accordingly, each elementary function is a holomorphic function on its natural domain which is an open subset of ${\mathbb C}$. In particular, the floor function $x\mapsto \lfloor x \rfloor, x\in {\mathbb R}$, is not regarded as a restriction of an elementary function in this answer.
I will not attempt to solve the problem where one uses discontinuous logarithmic function: Most likely, this can be done by considering multivalued holomorphic functions of several variables (single-valued holomorphic functions on Riemann domains over ${\mathbb C}^{k+1}$), but that would require substantially more work.
Every elementary function $f\in {\mathcal E}$ of the complex variable $z$ is defined by a formula $\varphi$ and a set of constants $c_1,..., c_k\in {\mathbb C}$. Treating the constants as independent complex variables $w_1,...,w_k$, we see that $\varphi$ defines a holomorphic function $F_{\varphi}=F(z,w_1,...,w_k)$ of $k+1$ complex variables defined on some open subset $\Omega$ of ${\mathbb C}^{n+1}$, its "natural domain." The original elementary function $f$ then is obtained by evaluating: $f(z)= F(z,c_1,...,c_k)$.
For most of the discussion below, I will fix $\varphi$; the discussion will only use the above description of $f$ in terms of a holomorphic function $F$ of several variables.
For each $n\in {\mathbb N}$, define $\Omega_n=\{\underline{w}=(w_1,...,w_k): (n,w_1,...,w_k)\in \Omega\}$.
Let $\Omega_\omega$ denote the infinite product
$$
\Omega_1\times \Omega_2\times ...,
$$
equipped with the product topology. I will use the notation $\Delta$ for the (small) diagonal in $\Omega_\omega$.
Then each $\overline{w}=(\underline{w}, \underline{w},....)\in \Delta$ defines the sequence $\theta(\overline{w}): n\mapsto F(n,w)$,
$$
\theta: \Delta \to Seq=Map({\mathbb N}, {\mathbb C}).$$
Remark. $\Omega_\omega, \Delta$ and $\theta$ of course depend on the original holomorphic function $F$.
I will equip the set of complex sequences $Seq$ with the following sup-metric:
$$
d(\eta,\zeta)= \sup_{n\in {\mathbb N}} \min(|\eta(n)-\zeta(n)|, 1).
$$
This metric is complete because the space $\ell_\infty$ is complete: If $d(\eta,\zeta)<1$ then the difference
$\eta-\zeta$ is a sequence in $\ell_\infty$.
Similarly to the map $\theta$ I will define maps $\theta_n: M_n \to Map(\{1,...,n\}, {\mathbb C})\cong {\mathbb C}^n$. Here $M_n$ is the (small) diagonal in the product $\Omega_1\times ...\times \Omega_n$; it is a complex manifold of dimension $k$.
The map $\theta_n$ sends each finite sequence
$$
\overline{w}=\underbrace{(\underline{w},..., \underline{w})}_{n~ \hbox{times}}
$$
to the map
$$
j\mapsto F(j, \underline{w}), j\in \{1,...,n\},
$$
$\underline{w}=(w_1,...,w_k)$.
Lemma 1. For each natural number $n > k$, the image of $\theta_n$ has measure zero in ${\mathbb C}^n$.
Proof. This is a very general fact: Since $M_n$ is a complex manifold of dimension $k<n$ and the map $\theta_n: M_n\to {\mathbb C}^n$ is $C^1$-smooth (actually, holomorphic), hence, its image has measure zero. qed.
Corollary 1. For each compact $K\subset \Omega_\omega$ the image $\theta(\Delta\cap K)$ is nowhere dense in $Seq$.
Proof. By Lemma 1, for each $n>k$ the image $\theta_n(K\cap \Delta\cap \Omega_1\times ...\times \Omega_n)$ is compact and nowhere dense in $Map(\{1,...,n\}, {\mathbb C})$. By compactness of $K$, for each sequence $\sigma\in cl(\theta(K\cap \Delta))\subset Seq$, the restriction of $\sigma$ to the integer interval $[1,n]$ lies in $\theta_n(K\cap \Delta)$. Hence, by Lemma 1, $\sigma|_{[1,n]}$ is the limit of a sequence
$$
\zeta_j\in Map(\{1,...,n\}, {\mathbb C}) \setminus \theta_n(K\cap \Delta\cap \Omega_1\times ...\times \Omega_n),$$
$$
\lim_{j\to\infty}\zeta_j= \sigma|_{[1,n]}.
$$
Extending each finite sequence $\zeta_j$ to the interval $[n+1,\infty)\cap {\mathbb N}$ by $\sigma|_{[n+1,\infty)}$ we obtain a sequence $\hat\zeta_j\in Seq$ which converges to $\sigma$ and, at the same time, $\hat\zeta_j\notin cl(\theta(K\cap \Delta))$. qed.
Corollary 2. $\theta(\Delta)$ is a union of countably many subsets of $Seq$ which are nowhere dense in $Seq$.
Proof. Since each $\Omega_n$ is an open subset of ${\mathbb C}^k$, there exists a sequence of compact subsets $K_j\subset \Delta\subset \Omega_\omega$ whose union is the entire $\Delta$, $j\in {\mathbb N}$. Thus, by Corollary 1, $\theta(\Delta)$ is the countable union of nowhere dense subsets
$$
\theta(K_j) \subset Seq.
$$
qed.
Now, back to elementary functions. There are only countably many formulae $\varphi$ defining elementary functions. Each formula $\varphi$ defines a holomorphic function of several variables $F_\varphi$. For each $F=F_\varphi$, by Corollary 2, the image of
$\theta=\theta_F: \Delta_F\to Seq$ is a countable union of nowhere dense subsets. Taking the union over all formulae $\varphi$, we conclude:
Corollary 3. The set of sequences of complex numbers defined by elementary functions is a countable union of nowhere dense subsets of $Seq$.
Thus, by the Baire Category Theorem, since $(Seq,d)$ is a complete metric space, we obtain:
Theorem. The set of sequences of complex numbers defined by elementary functions has empty interior in $Seq$.
