I'll prove the direction that 2 implies 1.

Using that $G$ is connected and 1-ended, you can obtain an increasing sequence of connected, finite diameter subgraphs
$$G'_0 \subset G'_1 \subset G'_2 \subset \cdots
$$
whose union is $G$, such that for all $i$, all but one component of the subgraph $G \setminus G'_i$ has finite diameter.

Since $G$ is locally finite, it follows that each $G'_i$ is finite, each finite diameter component of $G \setminus G'_i$ is finite, and there are only finitely many such components. Letting $G_i$ be the union of $G'_i$ with the finite diameter components of $G \setminus G'_i$, we obtain an increasing sequence of connected, **finite** subgraphs
$$G_0 \subset G_1 \subset G_2 \subset \cdots
$$
whose union is $G$, such that $G \setminus G_i$ consists of a single unbounded component.

Let $P_i = G_i \cap (G \setminus G_i)$, which is a finite set of points. For convenience, I'll pass to a subsequence so that $P_i$ is contained in the topological interior of $G_{i+1}$, hence $P_i$ is disjoint from $G \setminus G_{i+1}$.

Now I'll construct the desired infinite path $\gamma$, inductively constructing a strictly increasing family of initial segments
$$\gamma_0 \subset \gamma_1 \subset \gamma_2 \subset \cdots
$$
The path $\gamma_0$ starts anywhere, then runs around, concatenating edges, and hitting every vertex of $G_0$, ending at some $p_0 \in P_0$. Assuming by induction that $\gamma_i$ has been constructed to hit every point of $G_i$, ending at some $p_i \in P_i$, starting from $p_i$ the path $\gamma_{i+1} \setminus \gamma_i$ runs around in $G \setminus G_i$, concatenating edges, and hitting every point of $G_{i+1} \setminus G_i$, ending at some point $p_{i+1} \in P_{i+1}$.

The path $\gamma$ does hit every vertex, by construction. Also, clearly $\gamma_{i+1}$ hits each vertex of $G_i$ only finitely many times and $\gamma_{i+2} \setminus \gamma_{i+1}$ does not hit any vertex of $G_i$, hence $\gamma$ hits each vertex only finitely many times.