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, and we have
$$a=(A_j-\eta_j)/(2v_j), \quad j=1,2,$$
but for $b$ we obtain
$$av_j^2+bv_j+\eta_jv_j=\pi i,\quad i=1,2.$$
These two equations with one variable must be consistent,
which is very unlikely, certainly not for all $\lambda_1,\lambda_2$. To check this one has to compute $\eta_j$ for your lattices.