Yes, there are other spaces in which Borel's Theorem holds (property (BT)).
To see this, 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 balanced $0$-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.