Generally, one does not have a metric space embedded in some larger space where the "limits" of sequences that do not converge in $X$ may converge. So it really makes no sense to talk about "points to which the series converges but are not in $X$". As I noted in comments, if you really want to put (scare) quotes, they belong around the word "point", in so far as there is no such "point" to which these Cauchy sequences converge.
The solution to this is essentially the same as the one used to construct the reals from the rationals by considering all "possible limits" of Cauchy sequences: one constructs the completion of the space $X$ and embeds $X$ into that space. This completion, $Y$, is a complete metric space that comes equipped with an embedding $X\hookrightarrow Y$ such that (i) the embedding is uniformly continuous; (ii) [the image of] $X$ is dense in $Y$; and (iii) given any uniformly continuous function $f\colon X\to N$ into a complete metric space $N$, there is a unique uniformly continuous extension of $f$ to $\mathfrak{f}\colon Y\to N$. Viewing $X$ as a subspace of that $Y$, then you can talk about the limits of these Cauchy sequences in $X$ much like we can talk about the real limits of Cauchy sequences of rationals.
The universal property given above implies that if any such $Y$ exists, then it is unique up to a uniformly continuous homeomorphism, by the usual abstract nonsense arguments. So it suffices to construct any one such space. The standard construction mimics, as I mentioned above, the construction of $\mathbb{R}$ from $\mathbb{Q}$, specifically the construction of $\mathbb{R}$ as equivalence classes of rational Cauchy sequences. Namely, we let $C$ be the set of all Cauchy sequences of elements of $X$, and we define an equivalence relation on $C$ by letting $(x_n)\sim (y_n)$ if and only if $\lim\limits_{n\to\infty}d(x_n,y_n) = 0$. Then we let $Y$ be the quotient $C/\sim$, and define the metric by $D\left(\overline{(x_n)},\overline{(y_n)}\right) = \lim_{n\to\infty}d(x_n,y_n).$ One embeds $X$ into $Y$ by mapping $x$ to the class of the constant sequence $(x)$, and proves it has the appropriate properties.
Once you have this completion $Y$, now you can talk about those $X$-Cauchy sequences converging to points in $Y$ that are not in $X$; they converge in $Y$ because $Y$ is complete.