How do I prove that the Hadamard satisfies $Hequiv e^{ipi H/2}$?

Hint: Consider the series expansion of the exponential in the case of matrices $M$ that satisfy $M^2=-1$ (in your case $M=i frac{X+Z}{sqrt(2)}$. You should find something akin to Euler's formula which renders the proof trivial.

For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define $vec sigma = (sigma_1,sigma_2,sigma_3)$ to represent the three Pauli matrices $$sigma_1 = X = begin{bmatrix} 0 & 1 1 & 0 end{bmatrix}, ;;; sigma_2 = Y = begin{bmatrix} 0 & -i i & 0 end{bmatrix}, ;;; sigma_3 = Z = begin{bmatrix} 1 & 0 0 & -1 end{bmatrix}.$$ The Pauli matrices form a basis for the Lie algebra $mathfrak{su}_2$, and the corresponding Lie group elements, $U in SU(2)$, are given by the exponential map $$U = e^{i, vec phi , cdot vec sigma}, ;;; vec phi in mathbb{R}^3.$$ Define the vector $vec phi$ by a magnitude $alpha$ and unit vector $hat phi = (phi_1,phi_2,phi_3)$ such that $vec phi = alpha hat phi$. Simple multiplication shows that $(vec phi cdot vec sigma)^2 = alpha^2$. With this relationship, the Taylor expansion of $U$ works out very nicely. $$U = sum limits_{i=0}^infty frac{i^n}{n!} , (vec phi cdot vec sigma)^n = sum limits_{j=0}^infty frac{(-1)^j}{(2j)!} alpha^{2j} + i hat phi cdot vec sigma sum limits_{j=0}^infty frac{(-1)^j}{(2j + 1)!} , alpha^{2j+1}$$ $$=I , cos alpha + i hat phi cdot vec sigma sin alpha = begin{bmatrix} cos alpha + i phi_3 sin alpha && (phi_2 + i phi_1) sin alpha (-phi_2 + i phi_1) sin alpha && cos alpha - i phi_3 sin alpha end{bmatrix}.$$

With this formula it's simple to find the group element corresponding to given Lie algebra parameters. In the case of your specific question $alpha = tfrac{pi}{2}$ and $hat phi = (tfrac{1}{sqrt{2}}, 0, tfrac{1}{sqrt{2}})$. Plugging this in gives $$U_{x+z} = begin{bmatrix} frac{i}{sqrt{2}} && frac{i}{sqrt{2}} frac{i}{sqrt{2}} && -frac{i}{sqrt{2}} end{bmatrix} = frac{i(X + Z)}{sqrt{2}}.$$ In the underlying question from qiskit, $equiv$ is defined as equivalence modulo global phase, so, as desired, the result equals $tfrac{X+Z}{sqrt{2}}$ up to a global phase of $e^{ifrac{pi}{2}}$.

The more general question of determining what other Lie algebra parameterizations share this property (apart from trivial solutions, which are given by multiples of the identity) reduces to finding solutions to the equation $$e^{i vec phi , cdot , vec sigma} = e^{i theta} , hat phi cdot vec sigma ; (text{mod} ; theta).$$ This requires $cos alpha = 0$, which means vectors solving this equation will have $alpha = pm tfrac{pi}{2}$. From there it's relatively straightforward to see that solutions take the form of vectors with $alpha = pm frac{pi}{2}$ and $e^{pm i frac{pi}{2} hat phi , cdot , vec sigma} = pm i , hat phi cdot vec sigma$.

You can take the definition of the function $f(x)$ acting on any normal matrix to be such that if $$ M=sum_ilambda_i|lambda_iranglelanglelambda_i|, $$ then $$ f(M):=sum_if(lambda_i)|lambda_iranglelanglelambda_i|. $$ So for any $M$ such that $M^2=I$, then $M=P_+-P_-$ is described using projectors onto its $pm1$ eigenspaces, and $P_++P_-=I$. We can solve these simultaneously to get $P_+=(I+M)/2$ and $P_-=(I-M)/2$. Now, $$ f(M)=e^{it}P_++e^{-it}P_-=e^{it}(I+M)/2+e^{-it}(I-M)/2=cos(t)I+isin(t)M. $$