Solution:

If $m<2$ or $n<2$, then it is not possible to fit even one penny on the table, so the first player loses immediately. On the other hand, if $m \geq 2$ and $n \geq 2$, we will show that the first player can win. Let $O$ denote the center of the table, and let the first player place his first penny so that it is centered at $O$. The resulting configuration is centrally symmetric (i.e. it remains the same under a rotation of $180^{\circ}$ about $O$). Now, whenever the second player places a penny with some center $A$, the first player should respond with a penny centered at $B$, the $180^{\circ}$ rotation of $A$ about $O$. It is evident that the first player always leaves a centrally symmetric configuration; we must show that this move is actually legal.

Since placing the penny with center $A$ was legal, the symmetry of the previous configuration means that, before this penny was placed, a move centered at $B$ was also legal. So, after the second player's move at $A$, the first player can still legally move at $B$ unless the penny centered at $B$ intersects the penny centered at $A$. If $P$ is an intersection point, then $AP \leq 1$ and $PB \leq 1$, so $AB \leq 2$. But $O$ is the midpoint of $AB$, so $AO = AB / 2 \leq 1$. This means that $A$ lies inside the penny placed with center $O$; this is a contradiction. Hence, the move with center $B$ is, in fact, legal.

Thus, the first player can always legally move so as to preserve central symmetry. This move's availability means that the first player never loses. But since only a finite number of pennies can be placed without overlap, the game eventually ends, and it is the second player who loses.