Such triples exist, I think.
First, embed $PGL_2(p)$ in $S_{p+1}$ through its action on $1$-spaces from ${\mathbf F}_p^2$. This maps elements of order $p+1$ in $PGL_2(p)$ to $(p+1)$-cycles, and elements of order $p-1$ to $(p-1)$-cycles. In particular, every element of order $p+1$ or $p-1$ gets mapped to an odd permutation.
There are two conjugacy classes of elements of order two in $PGL_2(p)$. Elements from one class fix two $1$-spaces and elements from the other class fix none. So, elements of one class map to even permutations and elements of the other map to odd permutations under the given embedding.
It follows now that if $a,b \in PGL_2(p)$ have order two with $|ab| \in \{p-1,p+1\}$ then one of $a,b$ fixes two points and the other fixes none.
The stabilizer $B$ of a $1$-space in $PGL_2(p)$ is an extension of a cyclic group of order $p$ by a cyclic group of order $p-1$ that acts faithfully. It follows that if $a,b \in B$ are distinct elements of order two, then $|ab|=p$.
Now we are in good shape. Pick $g \in PGL_2(p)$ with $|g|=p+1$. Let $z=g^{(p+1)/2}$. As $g$ acts as a $(p+1)$-cycle, $z$ is a fixed-point-free involution. Now find involutions $a,b$, each with two fixed points, such that $|az|=p+1$ and $|bz|=p-1$. (This is certainly possible by the arguments above.) Now pick any $1$-space $V$ from ${\mathbf F}_p^2$. As $\langle g \rangle$ acts transitively on $1$-spaces, there exist a $\langle g \rangle$-conjugate $x$ of $a$ and a $\langle g \rangle$-conjugate $y$ of $b$ such that both $x$ and $y$ fix $V$.
As $g$ centralizes $z$, we get that $|xz|=p+1$ and $|yz|=p-1$. So, $x \neq y$. Now, as $x$ and $y$ fix a common $1$-space, $|xy|=p$.