If $T_h$ is the temperature of the hot reservoir and $T_c$ is the temperature of the cold reservoir, then the correct equation for the entropy balance on the working fluid is $$\Delta S=\frac{Q_h}{T_h}-\frac{Q_c}{T_c}+S_{gen}$$where $S_{gen}$ is the irreversible entropy generated during the cycle. But, since the engine is operating in a cycle and entropy is a function of state, $\Delta S=0$. So, $$\frac{Q_h}{T_h}-\frac{Q_c}{T_c}+S_{gen}=0\tag{1}$$. Also, since $S_{eng}$ is always greater or equal to zero, these results are likewise consistent with the Clausius inequality which, for this situation, becomes: $$\Delta S=0\gt\frac{Q_h}{T_h}-\frac{Q_c}{T_c}$$Since $W=Q_h-Q_c$, it follows from Eqn. 1 that:$$\frac{Q_h}{T_h}-\frac{Q_h-W}{T_c}+S_{gen}=0$$or equivalently,$$\eta=\frac{W}{Q_h}=\left(1-\frac{T_c}{T_h}\right)-\frac{S_{gen}T_c}{Q_h}\lt \left(1-\frac{T_c}{T_h}\right)$$So the efficiency is less than that of the reversible Carnot engine.

If $T_h$ is the temperature of the hot reservoir and $T_c$ is the temperature of the cold reservoir, then the correct equation for the entropy balance on the working fluid is $$\Delta S=\frac{Q_h}{T_h}-\frac{Q_c}{T_c}+S_{\text{gen}}$$where $S_{\text{gen}}$ is the irreversible entropy generated during the cycle. But, since the engine is operating in a cycle and entropy is a function of state, $\Delta S=0$. So, $$\frac{Q_h}{T_h}-\frac{Q_c}{T_c}+S_{\text{gen}}=0\tag{1}$$ Also, since $S_{\text{gen}}$ is always greater or equal to zero, these results are likewise consistent with the Clausius inequality which, for this situation, becomes: $$\Delta S=0\gt\frac{Q_h}{T_h}-\frac{Q_c}{T_c}$$Since $W=Q_h-Q_c$, it follows from Eqn. 1 that:$$\frac{Q_h}{T_h}-\frac{Q_h-W}{T_c}+S_{\text{gen}}=0$$or equivalently,$$\eta=\frac{W}{Q_h}=\left(1-\frac{T_c}{T_h}\right)-\frac{S_{\text{gen}}T_c}{Q_h}\lt \left(1-\frac{T_c}{T_h}\right)$$So the efficiency is less than that of the reversible Carnot engine.