3-dimensional matching variant
No, the variant is in P:


for each element $x_i \in X$
find the set $A_{x_i}$ of triples "reachable" from $x_i$:


start with all triples containing $x_i$ $A_{x_i} = \{ (x, y, z) \mid x = x_i \}$
iteratively add to $A_{x_i}$ all triples that share at least one element with all triples already in $A$

check if $E \setminus A_{x_i}$ is a solution and also for each triple $(x,y,z) \in E$ check if $E \setminus (A_{x_i} \setminus \{ (x,y,z) \})$ is a solution (remove at most one triple from $A_{x_i}$); if it's not a solution proceed with the next $x_i$ (step 1.)
if no solution is found, repeat the same algorithm 2–3 above "scanning" $y_i \in Y$ and checking $A_{y_i}$
if no solution is found, repeat the same algorithm 2–3 above "scanning" $z_i \in Z$ and checking $A_{z_i}$


To prove that  it is correct note that each $A_{x_i}$ contains at most one element of the solution $M$ of the 3DM:


by your variation constrait $E \setminus M$ is contained in at least one of the $A_{x_i},A_{y_i},A_{z_i}$, suppose $A_{x_i}$: $E \setminus M \subseteq A_{x_i}$;
furthermore $M \cap A_{x_i}$ must contain at most one triple from $A_{x_i}$, otherwise the 3DM constraint “covers each element of $X$, $Y$ and $Z$ exactly once” would be violated