Find all $a\in\mathbb{R}$ for which there exists a non-constant function $f: (0,1]\rightarrow\mathbb{R}$ such that \[a+f(x+y-xy)+f(x)f(y)\leq f(x)+f(y)\] for all $x,y\in(0,1].$
1. **Initial Condition Analysis:**
   We start by setting \( x = y = 1 \). The given inequality becomes:
   \[
   a + f(1) + f(1)^2 \leq f(1) + f(1)
   \]
   Simplifying, we get:
   \[
   a + f(1) + f(1)^2 \leq 2f(1)
   \]
   Rearranging terms, we have:
   \[
   a \leq 2f(1) - f(1) - f(1)^2 = f(1) - f(1)^2
   \]
   Let \( f(1) = c \). Then:
   \[
   a \leq c - c^2
   \]
   The function \( g(c) = c - c^2 \) reaches its maximum at \( c = \frac{1}{2} \), where \( g\left(\frac{1}{2}\right) = \frac{1}{4} \). Therefore:
   \[
   a \leq \frac{1}{4}
   \]

2. **Case \( a < \frac{1}{4} \):**
   For \( a < \frac{1}{4} \), we can choose \( f \) such that \( f: (0,1] \to \left[\frac{1}{2}, \min\left\{\frac{3}{4} - a, 1\right\}\right] \). We need to verify that the inequality holds:
   \[
   a + f(x + y - xy) + f(x)f(y) \leq f(x) + f(y)
   \]
   Rewriting the left-hand side:
   \[
   a + f(x + y - xy) + f(x)f(y) = a + f(x + y - xy) + (1 - f(x))(1 - f(y))
   \]
   Simplifying:
   \[
   a + f(x + y - xy) + 1 - f(x) - f(y) + f(x)f(y) = a + f(x + y - xy) + 1 - f(x) - f(y)
   \]
   Since \( f(x) \in \left[\frac{1}{2}, \min\left\{\frac{3}{4} - a, 1\right\}\right] \), we have:
   \[
   a + f(x + y - xy) + \frac{1}{4} - 1 \leq 0
   \]
   This holds because \( a < \frac{1}{4} \).

3. **Case \( a = \frac{1}{4} \):**
   For \( a = \frac{1}{4} \), we have:
   \[
   f(1) - f(1)^2 \leq \frac{1}{4}
   \]
   This implies:
   \[
   \left(f(1) - \frac{1}{2}\right)^2 \leq 0
   \]
   Therefore, \( f(1) = \frac{1}{2} \).

4. **Function Behavior for \( a = \frac{1}{4} \):**
   Setting \( y = 1 \) in the original inequality:
   \[
   \frac{1}{4} + f(x) + f(x)f(1) \leq f(x) + f(1)
   \]
   Since \( f(1) = \frac{1}{2} \):
   \[
   \frac{1}{4} + f(x) + \frac{1}{2}f(x) \leq f(x) + \frac{1}{2}
   \]
   Simplifying:
   \[
   \frac{1}{4} + \frac{3}{2}f(x) \leq f(x) + \frac{1}{2}
   \]
   Rearranging:
   \[
   \frac{1}{4} + \frac{1}{2}f(x) \leq \frac{1}{2}
   \]
   Therefore:
   \[
   \frac{1}{2}f(x) \leq \frac{1}{4}
   \]
   Hence:
   \[
   f(x) \leq \frac{1}{2}
   \]
   Since \( f(x) \geq \frac{1}{2} \) from the previous step, we conclude \( f(x) = \frac{1}{2} \) for all \( x \in (0,1] \).

5. **Contradiction:**
   Since \( f \) is constant, it contradicts the requirement that \( f \) is non-constant. Therefore, no such \( f \) exists for \( a = \frac{1}{4} \).

Conclusion:
\[
\boxed{a < \frac{1}{4}}
\]