The answer seems to be negative. Suppose that an entire function $f$ satisfies $f(z+v_i)=e^{A_iz}f(z)$, where $v_1$ and $v_2$ generate a lattice. Let $\Pi$ be the fundamental parallelogram of this lattice and integrate $f'/f$ over $\partial \Pi$. You obtain the ``Legendre's relation'': $$v_2A_1-v_1A_2=2\pi in,$$ where $n$ is the number of zeros of $f$ in $\Pi$.
Substituting your values, we see that $n=1$. Now $(f'/f)'$ is doubly periodic with respect to our lattice, having a single double pole per parallelogram. So we may assume (by shifting a pole to the origin) that $(f'/f)'=\wp+c,$ and two integrations integrations give $$f(z)=e^{P(z)}\sigma(z),$$ where $\sigma$ is the Weierstrass sigma function and $P$ is a polynomial of degree at most $2$. This is the general form of your $f$ (modulo a shift of the origin), if it exists.
Now let us try to find $P$. Sigma satisfies $$\sigma(z+v_j)=-e^{\eta_j(z+v_j)}\sigma(z),$$ where $\eta_j=\zeta(\omega_j)$, and $\zeta$ is the Weierstrass zeta function ($\zeta'=-\wp$), which gives $$P(z+v_j)=P(z)-\eta_j(z+v_j)+\pi i, \quad j=1,2.$$ Trying to find such a polynomial with your data, we just set $P(z)=az^2+bz+c$, and try to find $a,b,c$. Equation for $a$ is satisfied in view of Legendre's relation, but for $b$ we obtain $$v_j^2+bv_j+\eta_jv_j+\pi i=0,\quad i=1,2.$$ These two equations with one variable must be consistent, which is unlikely. To check this one has to compute $\eta_j$ for your lattice.