I think Thevenin (or Norton) equivalent circuits do not consider variable sources. The same refers to non-linear resistors (and other elements in AC scope). But I understand what you mean: you would like to have something like these.
In your case you should first select all the elements that are not dependent on other and do not alter other elements, and simplify them. The next step is to find all independent voltage/current sources.
Now combine non-linear static elements, like resistors. The combination of a linear object and a non-linear object is also non-linear object (but there is a theoretical possibility that two non-linear functions make a linear one).
At this moment you get: combined resistances that are (generally speaking) non-linear and do not alter anything and independent and dependent sources, and the elements that alter sources. If possible, combine independent sources.
That's the hardest task now: to combine independent sources with dependent. The Kirchoff's laws might be necessary here.
UPDATE
According to your circuit, this is not that difficult as it seems on the first sight. Please forgive me there are no exact calculations as I did them last time almost 20 years ago...
First of all, take a look at the non-ideal current source I1. Because it has R1 in parallel you can convert it to a non-ideal voltage source, which has resistance in series. This voltage source would have internal resistance 1 Ohm too and voltage R1 * 4Ix that is 4*Ix volts as R1 = 1 Ohm. I will name this new source as V2.
At the moment on the left side of the circuit you have non-ideal voltage source V2 (equivalent to I1 current source), its internal resistance (equivalent to R1), than voltage source V1. The R1 resistance is gone as it became internal load of voltage source. More reading about source transformation.
Because in the same branch there are two voltage sources you can combine them. So it is E = V1 + V2 which leads to (4 Ix - 10) V (- because V1 is in opposition to V2).
Now we have the first part of our task, the source. Now we're going to find equivalent resistance, and, moreover, we need to drive out Ix from source equation, because after combining resistances to one there will be no Ix.
As we know from Mr. Kirchoff, the load current (the one in R3), say I, divides in two: Ix and IL (IL flows through R3). The Ix is U2 / R2 and IL is U2 / (R3 + RL). You can write down proper equations yourself :).
Now you can find relation between Ix and IL (you need IL in equation of voltage source) and make E function of IL. If this source is no more function of Ix, you can combine other resistances to one equivalent. Do not forget source E internal resistance (the one driven from R1).
Please note that this method will lead you to have voltage source that is a function of load current (so in fact load resistance RL). This is normal as U2 depends on this load (that's why I've written at the very beginning it is not true Thevenin method).
When you say that the battery "draws a maximum of 10A" do you mean that when a true short circuit is applied to the battery, so that the terminal voltage is zero, that the supplied current is 10A? If so then your calculation is correct. If that is the maximum rated current you should draw from the battery then you need to know the voltage at the battery terminals when 10A is being drawn.
If you want to determine the Thevenin equivalent voltage and resistance without overloading the battery, then apply some known resistance \$R_L\$ and measure the output voltage as \$V_L\$. Measure the voltage without a load as \$V_{OC}\$. The voltage divider equation tells us that
\$ V_L = V_{OC} \frac{R_L}{R_{TH}\times R_L}\$
Solve for \$R_{TH}\$, and you know that \$V_{TH} = V_{OC}\$.
Best Answer
There are several ways to reduce a net of sources and resistors to its Thevenin equivalent. One way is like what you started to do. Find the open circuit voltage and the short circuit current.
Another way is to keep simplifying the circuit in steps. Take a source and two resistors and find its Thevenin equivalent on its own. Then replace the three parts with the two Thevenin equivalent parts in the circuit. Keep doing this until the circuit is a single voltage source in series with a single resistor.
However, in all cases you should start by drawing the circuit in a comprehensible way. Problems like this are often deliberately drawn to obscure the circuit. Never let that get in your way. Redraw the schematic properly with power voltages descending down the page, logical flow left to right, etc. That should be your first step here too.
You should always draw circuits properly before asking others to look at them. This web site is no exception.