I think I just answered my own question. No, this is not true. A positive answer to my question would imply, in particular, that two abelian extensions $L/K$ and $L'/K$ with the same conductor $\mathcal{C}$ are identical. However, this need not be the case.
For example, when $K=\mathbb{Q}$, and $L/\mathbb{Q}$ is abelian, and $m$ is the smallest positive integer such that $L\subseteq \mathbb{Q}(\zeta_{m})$, where $\zeta_m$ is a primitive $m$th root of unity, then the conductor $\mathcal{C}=C(L/\mathbb{Q})$ is given by the formula
$$C(L/\mathbb{Q})=\begin{cases}
m & \text{ if } L\subseteq \mathbb{R},\\
m\infty & \text{ otherwise}.
\end{cases}$$
Now take, for instance, $m=31$, and let $L_3$ and $L_5$ be, respectively, the unique extensions of $\mathbb{Q}$ contained in $\mathbb{Q}(\zeta_{31})$ of degree $3$ and $5$. Then, by the conductor formula, the conductor of $L_3$ and $L_5$ is $(31)$, because all their embeddings are real, and $L_3\not\subseteq L_5$ and $L_5\not\subseteq L_3$.
Note, however, that one direction is true: if $L/K$ and $L'/K$ are abelian extensions such that $K\subseteq L'\subseteq L$, then $\mathcal{C}'=C(L'/K)$ is a divisor of $\mathcal{C}=C(L/K)$. In order to show this, it is handy to remember that the conductor of an abelian extension $F/K$ is the greatest common divisor of all moduli $m$ such that $F\subseteq K_m$, where $K_m$ is the ray class field modulo $m$. In particular $L\subseteq K_{\mathcal{C}}$, and if $L'\subseteq L$, then $L'\subseteq K_{\mathcal{C}}$ as well. From this it follows that $\mathcal{C}'=C(L'/K)$ is a divisor of $\mathcal{C}$.
