Is it true that if $P(int_0^T f^2(s) ds<infty)=1$ then the exponential defines a density?

An easy counterexample is provided by the Bessel process (of dimension $2$). Let $X_t=|W_t|$, where $W$ is a two-dimensional Brownian motion started at some unit vector. Then, by Ito, $$ dX_t=frac{1}{2X_t},dt+dB_t,quad X_0=1, $$ for a one-dimensional BM $B$. Consider $f(t)=-frac{1}{2X_t}$ and note that this satisfies your assumption $P(int_0^1 f(u)^2,du<infty)=1$. If the stochastic exponential would define a density, then, by Girsanov, there were an equivalent probability measure $Q$ such that $(X_t)_{tin[0,1]}$ is a $Q$-BM. In particular, $$ Q(exists tin[0,1]:,X_t=0)>0, $$ but, by standard properties of two-dimensional BM, $$ P(exists tin[0,1]:,X_t=0)=0. $$ It is interesting to observe that, however, $$ mathbb{E}left[expleft(int_0^t f(s),ds-frac12int_0^t f(s)^2,dsright)right]<infty $$ for all $tin[0,1]$.