How can I identify that a restricted variant of Boolean SAT remains hard or not?
Your problem cannot be solved in polynomial time, for a boring reason: the size of the output might be exponential in the time of the input.  Therefore, there is no hope for a polynomial-time algorithm.

For instance, consider the boolean function $f:\{0,1\}^n \to \{0,1\}$ that outputs $1$ if its input vector has at least $n/2$ ones (i.e., its Hamming weight is at least $n/2$), and $0$ otherwise.  This function can be represented by a polynomial-size monotone formula.  However, there are exponentially many minimal $x$ such that $f(x)=1$.  In particular, each $x$ of Hamming weight $n/2$ is such a minimal input, and there are ${n \choose n/2}  \approx 2^n/\sqrt{n}$ such inputs.

So, there's no hope.

As Andrej Bauer points out, a related problem (testing whether there exists an input of Hamming weight at most $k$ that causes the boolean function to output $1$) is also NP-complete; see Prove NP-completeness of deciding satisfiability of monotone boolean formula.