Yes, there are other spaces in which Borel's Theorem holds (property (BT)).
Example 1. Note that every cartesian product of locally convex spaces with (BT) has (BT). In particular, any uncountable power ${\mathbb R}^I$ has (BT). Yet, ${\mathbb R}^I$ cannot be given a topology ${\mathcal O}$ which makes it a Fréchet space and is finer than the product topology ${\mathcal T}$.
In fact, for any such topology ${\mathcal O}$, we could choose a decreasing sequence of absolutely convex $0$-neighbourhoods $U_n$ in $({\mathbb R}^I,{\mathcal O})$ which form a basis of $0$-neighbourhoods. Let $C_n$ be the closure of $U_n$ in $({\mathbb R}^I,{\mathcal T})$.
Now $({\mathbb R}^I,{\mathcal T})$ is a Baire space (see Oxtoby, J.C., Cartesian products of Baire spaces, Fundam. Math. 49 (1961), 157-166).
As in the usual proof of the open mapping theorem, we see that each $C_n$ is a $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$. Since each $0$-neighbourhood in $({\mathbb R}^I,{\mathcal T})$ contains some $C_n$, we see that $({\mathbb R}^I,{\mathcal T})$ would be first countable, which is absurd.
This contradiction shows that ${\mathcal O}$ cannot exist.
Example 2. Recall that the usual proof of Borel's Theorem (for $V$ a Fréchet space) furnishes a smooth function $f\colon {\mathbb R}^d\to V$ of the form $f(x)=\sum_{\alpha \in{\mathbb N}_0^d} \;v_\alpha h(m_\alpha x)x^\alpha$, where $h\colon {\mathbb R}^d\to[0,1]$ is a cut-off function around $0$ and the $m_\alpha$ are positive constants. It can always be achieved that $m_\alpha\to\infty$ as $|\alpha|\to\infty$. As a consequence, the above sum is a finite sum (almost all summands vanish) for each fixed $x\in{\mathbb R}^d$, and likewise for all partial derivatives of the summands. It is clear from this observation that the completeness of $V$ does not play a role: The construction works just as well if $V$ is any (not necessarily complete) metrizable locally convex space.
Let $V$ be a metrizable locally convex space whose dimension as an abstract vector space is infinite and countable. By the preceeding, $V$ has (BT). Yet, there is no vector topology on $V$ making it a Fréchet space.
Let me mention that the product topology on ${\mathbb R}^I$ from Example 1 cannot be refined to a non-complete metrizable locally convex vector topology ${\mathcal O}$ either (by the same proof).