2 Fixed TeX

Let me explain why Beilinson's conjecture implies that $\iota$ is the zero map (thus your first question has conditionally a negative answer).

Let $\mathcal{E}$ be a proper regular model of $E$ over $\mathbf{Z}$. The morphism $E \to \mathcal{E}$ induces a $\mathbf{Q}$-linear map $\iota : K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)}$. The image of $\iota$ is the integral subspace $K_1(E)^{(2)}{\mathbf{Z}}$, K_1(E)^{(2)}_{\mathbf{Z}}$, which is also written $H^3{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$ in cohomological notations.

Now, what does Beilinson's conjecture predict for this group? We are concerned here with the motive $h^2(E)$, whose $L$-function is $L(H^2(E),s)=\zeta(s-1)$, and we are looking at the point $s=2$. Thus we are in the case of the "near central point" (see for example Schneider, Introduction to the Beilinson conjectures, \S Section 5, Conjecture II, or the articles by Beilinson and Nekovar mentioned in my comment).

Since the $L$-function has a pole, we have to introduce the group $N^1(E)=(\operatorname{Pic}(E)/\operatorname{Pic}^0(E)) \otimes \mathbf{Q}$ which is isomorphic to $\mathbf{Q}$ (more generally, the dimension should be equal to the order of the pole). There is a natural injective map $\psi : N^1(E) \to H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2))$. Then Beilinson's conjecture asserts that $r \oplus \psi$ induces an isomorphism

$$(H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2)) \otimes_{\mathbf{Q}} \mathbf{R}) \oplus (N^1(E) \otimes_{\mathbf{Q}} \mathbf{R}) \xrightarrow{\cong} H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2)).$$

Since the target space is $1$-dimensional and $N^1(E) \cong \mathbf{Q}$, this predicts in particular that $H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))=0$.

Moreover, it can be shown that the map $r$ is nonzero. As pointed out by profilesdroxford, the group $H^3_{\mathcal{M}}(E,\mathbf{Q}(2))$ is generated by symbols of the form $(P,\lambda)$ where $P$ is a closed point of $E$ and $\lambda \in \mathbf{Q}(P)^*$. I found a reference for this in Beilinson, Notes on absolute Hodge cohomology (Beilinson attributes this construction to Bloch and Quillen). Furthermore, in the same article the regulator of such elements is computed (in a more general setting). After some computations it turns out that $r([P,\lambda])$ is proportional to $\log | \operatorname{Nm}_{\mathbf{Q}(P)/\mathbf{Q}}(\lambda) |$. Thus $r$ is nonzero. Another useful reference is Dinakar Ramakrishnan's article on regulators.

It would be also interesting to compare the above construction with the construction proposed by profilesdroxford in his answer.

1

Let me explain why Beilinson's conjecture implies that $\iota$ is the zero map (thus your first question has conditionally a negative answer).

Let $\mathcal{E}$ be a proper regular model of $E$ over $\mathbf{Z}$. The morphism $E \to \mathcal{E}$ induces a $\mathbf{Q}$-linear map $\iota : K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)}$. The image of $\iota$ is the integral subspace $K_1(E)^{(2)}{\mathbf{Z}}$, which is also written $H^3{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$ in cohomological notations.

Now, what does Beilinson's conjecture predict for this group? We are concerned here with the motive $h^2(E)$, whose $L$-function is $L(H^2(E),s)=\zeta(s-1)$, and we are looking at the point $s=2$. Thus we are in the case of the "near central point" (see for example Schneider, Introduction to the Beilinson conjectures, \S 5, Conjecture II, or the articles by Beilinson and Nekovar mentioned in my comment).

Since the $L$-function has a pole, we have to introduce the group $N^1(E)=(\operatorname{Pic}(E)/\operatorname{Pic}^0(E)) \otimes \mathbf{Q}$ which is isomorphic to $\mathbf{Q}$ (more generally, the dimension should be equal to the order of the pole). There is a natural injective map $\psi : N^1(E) \to H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2))$. Then Beilinson's conjecture asserts that $r \oplus \psi$ induces an isomorphism

$$(H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2)) \otimes_{\mathbf{Q}} \mathbf{R}) \oplus (N^1(E) \otimes_{\mathbf{Q}} \mathbf{R}) \xrightarrow{\cong} H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2)).$$

Since the target space is $1$-dimensional and $N^1(E) \cong \mathbf{Q}$, this predicts in particular that $H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))=0$.

Moreover, it can be shown that the map $r$ is nonzero. As pointed out by profilesdroxford, the group $H^3_{\mathcal{M}}(E,\mathbf{Q}(2))$ is generated by symbols of the form $(P,\lambda)$ where $P$ is a closed point of $E$ and $\lambda \in \mathbf{Q}(P)^*$. I found a reference for this in Beilinson, Notes on absolute Hodge cohomology (Beilinson attributes this construction to Bloch and Quillen). Furthermore, in the same article the regulator of such elements is computed (in a more general setting). After some computations it turns out that $r([P,\lambda])$ is proportional to $\log | \operatorname{Nm}_{\mathbf{Q}(P)/\mathbf{Q}}(\lambda) |$. Thus $r$ is nonzero. Another useful reference is Dinakar Ramakrishnan's article on regulators.

It would be also interesting to compare the above construction with the construction proposed by profilesdroxford in his answer.