$\def\ddt{\frac{d}{dt}}\def\l\{\left}\def\r{\right}\def\rmS{S}$The network is over-idealized. Any inductor currents $i_1$, $i_2$ with $i_1+i_2 = \frac{e}{R}$ are possible at steady state.
If you are given initial currents $i_1(0),i_2(0)$ then current over time results with $i_\rmS:=i_1+i_2$ from \begin{align} \ddt i_1(t) &= \frac1{L_1}\l(e - R\cdot (i_1(t)+i_2(t))\r)\\ \ddt i_2(t) &= \frac1{L_2}\l(e - R\cdot (i_1(t)+i_2(t))\r) \end{align}\begin{align} \ddt i_1(t) &= \frac1{L_1}\l(e - R\cdot i_\rmS(t)\r)\\ \ddt i_2(t) &= \frac1{L_2}\l(e - R\cdot i_\rmS(t)\r) \end{align} Adding the equations gives \begin{align} \ddt i_\rmS (t) = \l(\frac1{L_1}+\frac1{L_2}\r)\l(e - R\cdot i_\rmS\r). \end{align} with $i_\rmS:=i_1+i_2$. NowNow you can determine the sum current $i_\rmS(t)$ over time with the initial condition $i_\rmS(0)=i_1(0)+i_2(0)$. From that you can determine $e-R\cdot i_\rmS$ and therefore $i_1(t)$ and $i_2(t)$. The limit $t\rightarrow\infty$ gives you the steady state values.
The situation changes if the inductors have internal resistance (which is not indicated in the circuit diagram). Then you can calculate the steady-state currents with the inductors replaced by the internal resistances.