We show that if $\alpha > 2$ then Hephaestus wins, but when $\alpha = 2$ (and hence $\alpha \le 2$) Hephaestus cannot contain even a single-cell flood initially.

Strategy for $\alpha > 2$: Impose $\mathbb{Z}^2$ coordinates on the cells. Adding more flooded cells does not make our task easier, so let us assume that initially the cells $(x, y)$ with $|x| + |y| \le d$ are flooded for some $d \ge 2$; thus on Hephaestus's $k$th turn, the water is contained in $|x| + |y| \le d + k - 1$. Our goal is to contain the flood with a large rectangle.
We pick large integers $N_1$ and $N_2$ such that
$$\alpha N_1 > 2N_1 + (2d + 3) \\ \alpha(N_1 + N_2) > 2N_2 + (6N_1 + 8d + 4).$$
Mark the points $X_i$, $Y_i$ as shown in the figure for $1 \le i \le 6$. The red figures indicate the distance between the marked points on the rectangle.

We follow the following plan.
* Turn 1: place wall $X_1Y_1$. This cuts off the flood to the north.
* Turns 2 through $N_1 + 1$: extend the levee to segment $X_2Y_2$. This prevents further flooding to the north.
* Turn $N_1 + 2$: add in broken lines $X_4X_3X_2$ and $Y_4Y_3Y_2$ all at once. This cuts off the flood west and east.
* Turns $N_1 + 2$ to $N_1 + N_2 + 1$: extend the levee along segments $X_4X_5$ and $Y_4Y_5$. This prevents further flooding west and east.
* Turn $N_1 + N_2 + 2$: add in the broken line $X_5X_6Y_6Y_5$ all at once and win.

Proof for $\alpha = 2$: Suppose Hephaestus contains the flood on his $(n+1)$st turn. We prove that $\alpha > 2$ by showing that in fact at least $2n + 4$ walls have been constructed.
Let $c_0, c_1, \dots, c_n$ be a path of cells such that $c_0$ is the initial cell flooded, and in general $c_n$ is flooded on Poseidon's $n$th turn from $c_{n-1}$. The levee now forms a closed loop enclosing all $c_i$.

Claim — If $c_i$ and $c_j$ are adjacent then $|i - j| = 1$.
Proof. Assume $c_i$ and $c_j$ are adjacent but $|i - j| > 1$. Then the two cells must be separated by a wall. But the levee forms a closed loop, and now $c_i$ and $c_j$ are on opposite sides. $\square$

Thus the $c_i$ actually form a path. We color green any edge of the unit grid (wall or not) which is an edge of exactly one $c_i$ (i.e. the boundary of the polyomino). It is easy to see there are exactly $2n + 4$ green edges.

Now, from the center of each cell $c_i$, shine a laser towards each green edge of $c_i$ (hence a total of $2n + 4$ lasers are emitted). An example below is shown for $n = 6$, with the levee marked in brown.

Claim — No wall is hit by more than one laser.
Proof. Assume for contradiction that a wall $w$ is hit by lasers from $c_i$ and $c_j$. WLOG that laser is vertical, so $c_i$ and $c_j$ are in the same column (e.g. $(i, j) = (0, 5)$ in figure). We consider two cases on the position of $w$.
* If $w$ is between $c_i$ and $c_j$, then we have found a segment intersecting the levee exactly once. But the endpoints of the segment lie inside the levee. This contradicts the assumption that the levee is a closed loop.
* Suppose $w$ lies above both $c_i$ and $c_j$ and assume WLOG $i < j$. Then we have found that there is no levee at all between $c_i$ and $c_j$.
Let $\rho \ge 1$ be the distance between the centers of $c_i$ and $c_j$. Then $c_j$ is flooded in a straight line from $c_i$ within $\rho$ turns, and this is the unique shortest possible path. So this situation can only occur if $j = i + \rho$ and $c_i, \dots, c_j$ form a column. But then no vertical lasers from $c_i$ and $c_j$ may point in the same direction, contradiction.

Since neither case is possible, the proof ends here. $\Box$

This implies the levee has at least $2n + 4$ walls (the number of lasers) on Hephaestus's $(n+1)$st turn. So $\alpha \ge \frac{2n+4}{n+1} > 2$.