(RESTART) Here is a complete solution.

Call a graph *3-edge-connected* if the number of components cannot be increased by removing only 1 or 2 edges. (This allows for the graph to be disconnected already.)

Let $R~$ be a ring that has an identity and a left inverse of 2, which I'll call $\frac12$. (For example, a field of characteristic other than 2.)

Given a graph $G~$ with edge-set $E$, let $RE~$ be the module of formal sums $\sum_{e\in E} c_e e$, where each $c_e\in R$, with the usual operations of addition, and multiplication on the left by field elements. Let $RC~$ be the submodule generated by the (edge sets of) cycles.

**Theorem A.** $RE=RC~$ iff $G$ is 3-edge-connected.

**Proof.** If $e$ is an edge cut by itself (i.e., a bridge), it lies on no cycles, so $e\notin RC$. If $e,e'$ is an edge cut of size 2, then any cycle containing $e$ must also contain $e'$, so again $e\notin RC$. Therefore 3-edge-connectivity is necessary.

Conversely, assume $G~$ is 3-edge-connected and choose any $e=(v,w)\in E$. The graph after removing $e$, call it $G-e$, is 2-edge-connected, so by Menger's Theorem there are two edge-disjoint paths $P_1,P_2$ in $G-e$ from $v$ to $w$. Now note that
$$ e = \tfrac12((P_1\cup \{e\}) + (P_2\cup \{e\}) - (P_1\cup P_2)). $$
The three terms on the right are in $RC$, the first two because they are cycles and the third because the union of two edge-disjoint paths with the same endpoints is an edge-disjoint union of cycles. Therefore $e\in RC$.

(ADDITION) Let $u$ be a vertex, and let $C_u$ be the set of cycles passing through u.

**Theorem B.** If $G~$ is 3-connected, then $RC_u=RE$.

**Proof.** Let $C~$ be a cycle that does *not* pass through $u$. Since $G~$ is 3-connected, there are 3 paths from $u~$ to $C~$ which are vertex disjoint except at $u$. This divides $C~$ into 3 segments.
Now add the 3 evident cycles through $u~$ that use two segments of $C~$ and subtract the 3 evident cycles through $u~$ that use one segment of $C$. The result is $C$. That is, we can use the cycles through $u~$ to get all the cycles.
Therefore, $RC_u=RC$, and now apply Theorem A.

If we take two 3-connnected graphs, and glue them together at a single vertex $u$, then also $RC_u=RE$ for this graph. Yet it isn't 3-connected. So the "if" in Theorem B can't be changed to "iff". We can make an incomplete converse though.

**Theorem C.** If $G~$ is a connected graph and $RC_u=RE$, then $u$ lies on every 1-cut or 2-cut.

**Proof.** If there is a 1-cut other than $u$, clearly cycles through $u~$ can't reach the edges on the other side of the cut, so $RC_u\ne RE$. Now suppose there is a 2-cut
$\{x,y\}$ neither of which is $u$. Let $E_x$ be the edges of $G~$ which are incident to $x~$ but can't be reached from $u~$ except via the 2-cut. Similarly $E_y$. Every cycle through $u~$ either uses none of $E_x$ or $E_y$, or uses exactly one edge from each. Therefore, every vector in $RC_u$ has the same total coefficient for $E_x$ as it does for $E_y$, which is a linear dependence implying $RC_u\ne RE$.