How would one in general go about finding some polynomial $\Pi$ defined on say $[a, b]$ such that
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Let's ignore the no-zeroes-in-between requirement first.
First, choose an arbitrary polynomial with sufficiently many free parameters, for example $\Pi(x) = rx^3+sx^2+tx+u$, or for this case more cleverly $\Pi(x) = (rx+s)(x-a)(x-b)$.
Now let's look at the equations $\Pi(a) = 0, \Pi(b) = 0, \Pi'(a) = d_a, \Pi'(b) = d_b$. If we evaluate $\Pi$ there, we'll get a system of linear equations in $r, s, t$ and $u$. This should be easy to solve. You will get some $\Pi$ such that the boundary equations are satisfied.
Next, you'll want to check whether the no-zeroes requirement is satisfied. For the case $\text{sgn}(d_a) \neq \text{sgn}(d_b)$, this should already suffice. For the case $d_ad_b > 0$, this isn't satisfied, and it never will be, because any continuous function satisfying those boundary conditions will have a zero in $(a, b)$.
This leaves the case $d_a = d_b = 0$. Using the method above, you will get $\Pi = 0$, which doesn't suit your constraints. In this case you can try using more free variables to get a whole set of solutions to your linear equation system, for example by adding a $qx^4$ into the term or something. This will yield more solutions which are better suited to fit your constraints.
Or you could be really clever and simply manually choose the polynomial $(x-a)^2(x-b)^2$ to satisfy your requirements.
You can find cubic polynomials $f$ and $g$ that are zero at $a$ and $b$, are positive on $(a,b)$ and such that $f'(a) = 1$, $f'(b) = 0$, $g'(a) = 0$, and $g'(b) = 1$. (Hint: $f$ is a multiple of $(x-a)(x-b)^2$.) Then $\Pi = d_a f - d_b g$ will do the job (since $d_a$ and $d_b$ must have opposite signs if they are both nonzero).
If $d_a = d_b = 0$ and you don't consider $\Pi = 0$ to be valid, then you can use a quartic polynomial like $(x-a)^2(x-b)^2$.