Here is how we can get $\approx (mn+2m+2n)/5$ tiles. Place the numbers $1$ through $5$ in each square as shown below:
$$
\begin{array}{ccccccc}
1&4&2&5&3&1&\cdots\\
2&5&3&1&4&2&\\
3&1&4&2&5&3&\\
4&2&5&3&1&4&\\
5&3&1&4&2&5&\cdots\\
\vdots&&&&&\vdots&\ddots
\end{array}
$$
In words: the columns read $1,2,3,4,5$ periodically as you go down, and the pattern shifts down by 2 as you move right.
Now, color all $1$ tiles black. Then no more black tiles can be placed in the interior, because all numbers are neighbored by a $1$. The only place this breaks down is at the border: normally, a $2$ is covered by the $1$ tile above it, but $2$'s on the top border aren't covered. Same thing happens for the right, bottom and left border, with the numbers $3$, $5$ and $4$ respectively. Placing a black tile on each of these bad cases means no more can be placed.
Since there are about $\frac{mn}{5}$ ones, and about $\frac{m}{5}$ bad numbers on the left and right borders, and about $\frac{n}{5}$ on the top and bottom, my formula is as claimed.
This could be optimized further by initially coloring all of a different number black, instead of one, in order to minimize the bad cases.